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NCERT Exemplar for Class 12 Maths Chapter-7 (Book Solutions)

NCERT Exemplar for Class 12 Maths Chapter-7 (Book Solutions)

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Free PDF download of NCERT Exemplar for Class 12 Maths Chapter 7 - Integrals solved by expert Maths teachers on Vedantu.com as per NCERT (Central Board of Secondary Education) Book guidelines. All Chapter 7 - Integrals Exercise questions with solutions to help you to revise the complete syllabus and score more marks in your Examinations.

NCERT Exemplar for Class 12 Maths, Chapter 7 - Integrals is available on Vedantu and its solutions are available on Vedantu in PDF form. The PDF of Chapter 7 - Integrals consists of important topics and additional questions.

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Access NCERT Exemplar Solutions for CBSE Class 12 Mathematics Chapter 7 - Integrals

Solved Examples

Short Answer (S.A.)

1. Integrate $(\frac{2 a}{\sqrt{x}} - \frac{b}{x^{2}} + 3 c \sqrt[3]{x^{2}})$ w.r.t. $x .$

Ans: let $I = \int (\frac{2 a}{\sqrt{x}} - \frac{b}{x^{2}} + 3 c \sqrt[3]{x^{2}}) d x$

⇒ $I = 2 a \int x^{- \frac{1}{2}} d x - b \int x^{- 2} d x + 3 c \int x^{\frac{2}{3}} d x$

⇒ $I = 2 a \times 2 \times \sqrt{x} + \frac{b}{x} + \frac{9}{5} c x^{\frac{5}{3}} + C$

⇒ $I = 4 a \sqrt{x} + \frac{b}{x} + \frac{9}{5} c x^{\frac{5}{3}} + C$

Hence $\int (\frac{2 a}{\sqrt{x}} - \frac{b}{x^{2}} + 3 c \sqrt[3]{x^{2}}) d x = 4 a \sqrt{x} + \frac{b}{x} + \frac{9}{5} c x^{\frac{5}{3}} + C .$

2. Evaluate $\int \frac{3 a x}{b^{2} + c^{2} x^{2}} d x$

Ans: let $I = \int \frac{3 a x}{b^{2} + c^{2} x^{2}} d x$

Now, put $b^{2} + c^{2} x^{2} = t$

⇒ $2 c^{2} x d x = d t$

⇒ $x d x = \frac{d t}{2 c^{2}}$

⇒ $3 a x d x = \frac{3 a d t}{2 c^{2}}$

Therefore,

⇒ $I = \frac{3 a}{2 c^{2}} \int \frac{d t}{t}$

⇒ $I = \frac{3 a}{2 c^{2}} \log | t | + C$

⇒ $I = \frac{3 a}{2 c^{2}} \log | b^{2} + c^{2} x^{2} | + C$

Hence, $\int \frac{3 a x}{b^{2} + c^{2} x^{2}} d x = \frac{3 a}{2 c^{2}} \log | b^{2} + c^{2} x^{2} | + C .$

3. Verify the following using the concept of integration as an antiderivative.

$\int \frac{x^{3}}{x + 1} d x = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log | x + 1 | + C$

Ans: here, L.H.S $= x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log | x + 1 | + C$

Now, differentiate w.r.t $x$ , we get

⇒ L.H.S $= \frac{d}{d x} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log | x + 1 | + C)$

$= 1 - \frac{2 x}{2} + \frac{3 x^{2}}{3} - \frac{1}{(x + 1)} + 0$

$= 1 - x + x^{2} - \frac{1}{(x + 1)}$

$= \frac{(x + 1) - x (x + 1) + x^{2} (x + 1) - 1}{(x + 1)}$

$= \frac{x + 1 - x^{2} - x + x^{3} + x^{2} - 1}{(x + 1)}$

$= \frac{x^{3}}{(x + 1)}$

Therefore, $(x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log | x + 1 | + C) = \int \frac{x^{3}}{(x + 1)} d x$

Hence verified.

4. Evaluate $\int \sqrt{\frac{1 + x}{1 - x}} d x,$ $x \neq 1$ .

Ans: Let $I = \int \sqrt{\frac{1 + x}{1 - x}} d x$

⇒ $I = \int \sqrt{\frac{(1 + x) (1 + x)}{(1 - x) (1 + x)}} d x$

⇒ $I = \int \sqrt{\frac{{(1 + x)}^{2}}{1 - x^{2}}} d x$

⇒ $I = \int \frac{1 + x}{\sqrt{1 - x^{2}}} d x$

⇒ $I = \int \frac{1}{\sqrt{1 - x^{2}}} d x + \int \frac{x}{\sqrt{1 - x^{2}}} d x$

⇒ $I = s i n^{- 1} x + \int \frac{x}{\sqrt{1 - x^{2}}} d x$

Now, put $1 - x^{2} = t^{2}$

⇒ $- 2 x d x = 2 t d t$

⇒ $x d x = - t d t$

Therefore,

⇒ $I = s i n^{- 1} x - \int \frac{t d t}{t}$

⇒ $I = s i n^{- 1} x - \int d t$

⇒ $I = s i n^{- 1} x - t + C$

⇒ $I = s i n^{- 1} x - \sqrt{1 - x^{2}} + C$

Hence, $\int \sqrt{\frac{1 + x}{1 - x}} d x = s i n^{- 1} x - \sqrt{1 - x^{2}} + C$

5. Evaluate $\int \frac{d x}{\sqrt{(x - α) (β - x)}}$ , $β > α$

Ans: Let $I = \int \frac{d x}{\sqrt{(x - α) (β - x)}}$

Now, put $x - α = t^{2}$

⇒ $d x = 2 t d t$

And $β - x = β - (t^{2} + α) = β - α - t^{2}$

Therefore,

⇒ $I = \int \frac{2 t d t}{\sqrt{t^{2} (β - α - t^{2})}}$

⇒ $I = 2 \int \frac{d t}{\sqrt{(β - α - t^{2})}}$

⇒ $I = 2 \int \frac{d t}{\sqrt{[{(\sqrt{β - α})}^{2} - t^{2}]}}$

⇒ $I = 2 s i n^{- 1} \frac{t}{\sqrt{β - α}} + C$

⇒ $I = 2 s i n^{- 1} \frac{\sqrt{x - α}}{\sqrt{β - α}} + C$

Hence, $\int \frac{d x}{\sqrt{(x - α) (β - x)}} = 2 s i n^{- 1} \frac{\sqrt{x - α}}{\sqrt{β - α}} + C .$

6. Evaluate $\int t a n^{8} x s e c^{4} x d x$

Ans: let $I = \int t a n^{8} x s e c^{4} x d x$

⇒ $I = \int t a n^{8} x (s e c^{2} x) s e c^{2} x d x$

⇒ $I = \int t a n^{8} x (1 + t a n^{2} x) s e c^{2} x d x$

Now, put $\tan x = t$

⇒ $s e c^{2} x d x = d t$

Therefore,

⇒ $I = \int t^{8} (1 + t^{2}) d t$

⇒ $I = \int (t^{8} + t^{10}) d t$

⇒ $I = \frac{t^{9}}{9} + \frac{t^{11}}{11} + C$

⇒ $I = \frac{t a n^{9} x}{9} + \frac{t a n^{11} x}{11} + C$

Hence, $\int t a n^{8} x s e c^{4} x d x = \frac{t a n^{9} x}{9} + \frac{t a n^{11} x}{11} + C$

7. Find $\int \frac{x^{3}}{x^{4} + 3 x^{2} + 2} d x$

Ans: Let $I = \int \frac{x^{3}}{x^{4} + 3 x^{2} + 2} d x$

⇒ $I = \int \frac{x^{2} . x d x}{x^{4} + 3 x^{2} + 2}$

Now, put $x^{2} = t$

⇒ $2 x d x = d t$

⇒ $x d x = \frac{d t}{2}$

Therefore,

⇒ $I = \frac{1}{2} \int \frac{t d t}{t^{2} + 3 t + 2}$

⇒ $I = \frac{1}{2} \int \frac{t d t}{(t + 1) (t + 2)}$

Consider $\frac{t}{(t + 1) (t + 2)} = \frac{A}{t + 1} + \frac{B}{t + 2}$

⇒ $\frac{t}{(t + 1) (t + 2)} = \frac{A t + 2 A + B t + B}{(t + 1) (t + 2)}$

⇒ $\frac{t}{(t + 1) (t + 2)} = \frac{(A + B) t + 2 A + B}{(t + 1) (t + 2)}$

⇒ $t = (A + B) t + 2 A + B$

On comparing both sides, we get

⇒ $A + B = 1$ and $2 A + B = 0$

On solving, we get

⇒ $A = - 1, B = 2$

Therefore,

⇒ $I = \frac{1}{2} \int \frac{t d t}{(t + 1) (t + 2)}$

⇒ $I = \frac{1}{2} \int [\frac{- 1}{t + 1} + \frac{2}{t + 2}] d t$

⇒ $I = \frac{1}{2} (- \log | t + 1 | + 2 \log | t + 2 |) + C$

⇒ $I = - \frac{1}{2} \log | t + 1 | + \log | t + 2 | + C$

⇒ $I = - \log | \sqrt{t + 1} | + \log | t + 2 | + C$

⇒ $I = \log | \frac{t + 2}{\sqrt{t + 1}} | + C$

⇒ $I = \log | \frac{x^{2} + 2}{\sqrt{x^{2} + 1}} | + C$

Hence, $I = \int \frac{x^{3}}{x^{4} + 3 x^{2} + 2} d x = \log | \frac{x^{2} + 2}{\sqrt{x^{2} + 1}} | + C$

8. Find $\int \frac{d x}{2 s i n^{2} x + 5 c o s^{2} x}$

Ans: Let $I = \int \frac{d x}{2 s i n^{2} x + 5 c o s^{2} x}$

Dividing numerator and denominator by $c o s^{2} x,$ we get

⇒ $I = \int \frac{s e c^{2} x d x}{2 t a n^{2} x + 5}$

Put $\tan x = t$

⇒ $s e c^{2} x d x = d t$

Therefore,

⇒ $I = \int \frac{d t}{2 t^{2} + 5}$

⇒ $I = \frac{1}{2} \int \frac{d t}{t^{2} + \frac{5}{2}}$

⇒ $I = \frac{1}{2} \int \frac{d t}{t^{2} + {(\sqrt{\frac{5}{2}})}^{2}}$

⇒ $I = \frac{1}{2} \times \sqrt{\frac{2}{5}} t a n^{- 1} \frac{\sqrt{2} t}{\sqrt{5}} + C$

⇒ $I = \frac{1}{\sqrt{10}} t a n^{- 1} (\frac{\sqrt{2} t a n x}{\sqrt{5}}) + C$

Hence, $\int \frac{d x}{2 s i n^{2} x + 5 c o s^{2} x} = \frac{1}{\sqrt{10}} t a n^{- 1} (\frac{\sqrt{2} t a n x}{\sqrt{5}}) + C .$

9. Evaluate $\underset{- 1}{\int^{2}} (7 x - 5) d x$ as a limit of sums.

Ans: here, we have $a = - 1$ , $b = 2$ and $h = \frac{2 + 1}{n} \Rightarrow n h = 3$

And $f (x) = 7 x - 5$

Now, we have

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = lim_{h \to 0} h [f (- 1) + f (- 1 + h) + f (- 1 + 2 h) \dots + f (- 1 + (n - 1) h]$

Also we have,

⇒ $f (- 1) = - 7 - 5 = - 12$

⇒ $f (- 1 + h) = - 7 + 7 h - 5 = - 12 + 7 h$

⇒ $f (- 1 + (n - 1) h) = 7 (n - 1) h - 12 =$

Therefore,

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = lim_{h \to 0} h [- 12 + (- 7 h - 12) + (- 14 h - 12) \dots + (7 (n - 1) h - 12]$

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = lim_{h \to 0} h [7 h [1 + 2 + \dots + (n - 10)] - 12 n]$

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = lim_{h \to 0} h [7 h \times \frac{n (n - 1)}{2} - 12 n]$

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = lim_{h \to 0} h [\frac{7}{2} (n h) (n h - h) - 12 n h]$

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = \frac{7}{2} \times 3 \times 3 - 12 \times 3$

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = \frac{63}{2} - 36$

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = \frac{63 - 72}{2}$

⇒ $\underset{- 1}{\int^{2}} (7 x - 5) d x = \frac{- 9}{2}$

Hence, $\underset{- 1}{\int^{2}} (7 x - 5) d x = \frac{- 9}{2} .$

10. Evaluate $\underset{0}{\int^{\frac{π}{2}}} \frac{t a n^{7} x}{c o t^{7} x + t a n^{7} x} d x$

Ans: Let $I = \underset{0}{\int^{\frac{π}{2}}} \frac{t a n^{7} x}{c o t^{7} x + t a n^{7} x} d x$ …………. (i)

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{t a n^{7} (\frac{π}{2} - x)}{c o t^{7} (\frac{π}{2} - x) + t a n^{7} (\frac{π}{2} - x)} d x$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{c o t^{7} x}{t a n^{7} x + c o t^{7} x} d x$ …………….. (ii)

Now, adding eq (i) and (ii), we get

⇒ $2 I = \underset{0}{\int^{\frac{π}{2}}} \frac{t a n^{7} x + c o t^{7} x}{t a n^{7} x + c o t^{7} x} d x$

⇒ $2 I = \underset{0}{\int^{\frac{π}{2}}} d x$

⇒ $2 I = {[x]}_{0}^{\frac{π}{2}}$

⇒ $2 I = \frac{π}{2}$

⇒ $I = \frac{π}{4}$

Hence, $\underset{0}{\int^{\frac{π}{2}}} \frac{t a n^{7} x}{c o t^{7} x + t a n^{7} x} d x = \frac{π}{4} .$

11. Find $\underset{2}{\int^{8}} \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}} d x$

Ans: Let $I = \underset{2}{\int^{8}} \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}} d x$ ………….. (i)

⇒ $I = \underset{2}{\int^{8}} \frac{\sqrt{8 + 2 - (10 - x)}}{\sqrt{8 + 2 - x} + \sqrt{8 + 2 - (10 - x)}} d x$

⇒ $I = \underset{2}{\int^{8}} \frac{\sqrt{10 - 10 + x}}{\sqrt{10 - x} + \sqrt{10 - 10 + x}} d x$

⇒ $I = \underset{2}{\int^{8}} \frac{\sqrt{x}}{\sqrt{10 - x} + \sqrt{x}} d x$ ……………… (ii)

Now, adding eq (i) and (ii), we get

⇒ $2 I = \underset{2}{\int^{8}} \frac{\sqrt{10 - x} + \sqrt{x}}{\sqrt{10 - x} + \sqrt{x}} d x$

⇒ $2 I = \underset{2}{\int^{8}} d x$

⇒ $2 I = {[x]}_{2}^{8}$

⇒ $2 I = 8 - 2$

⇒ $2 I = 6$

⇒ $I = 3$

Hence, $\underset{2}{\int^{8}} \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}} d x = 3.$

12. Find $\underset{0}{\int^{\frac{π}{4}}} \sqrt{1 + \sin 2 x} d x$

Ans: Let $I = \underset{0}{\int^{\frac{π}{4}}} \sqrt{1 + \sin 2 x} d x$

⇒ $I = \underset{0}{\int^{\frac{π}{4}}} \sqrt{s i n^{2} x + c o s^{2} x + 2 s i n x . c o s x} d x$

⇒ $I = \underset{0}{\int^{\frac{π}{4}}} \sqrt{{(\sin x + \cos x)}^{2}} d x$

⇒ $I = \underset{0}{\int^{\frac{π}{4}}} (\sin x + \cos x) d x$

⇒ $I = {[- \cos x + \sin x]}_{0}^{\frac{π}{4}}$

⇒ $I = [- \cos \frac{π}{4} + \sin \frac{π}{4} - (- \cos 0 + \sin 0)]$

⇒ $I = [- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - (- 1 + 0)]$

⇒ $I = 1$

Hence, $\underset{0}{\int^{\frac{π}{4}}} \sqrt{1 + \sin 2 x} d x = 1.$

13. Find $\int x^{2} t a n^{- 1} x d x$

Ans: Let $I = \int x^{2} t a n^{- 1} x d x$

⇒ $I = t a n^{- 1} x \int x^{2} d x - \int (\frac{d}{d x} (t a n^{- 1} x) \int x^{2} d x) d x$

⇒ $I = \frac{x^{3}}{3} t a n^{- 1} x - \int \frac{1}{1 + x^{2}} . \frac{x^{3}}{3} d x$

⇒ $I = \frac{x^{3}}{3} t a n^{- 1} x - \frac{1}{3} \int \frac{x^{3}}{1 + x^{2}} d x$

⇒ $I = \frac{x^{3}}{3} t a n^{- 1} x - \frac{1}{3} \int (x - \frac{x}{1 + x^{2}}) d x$

⇒ $I = \frac{x^{3}}{3} t a n^{- 1} x - \frac{1}{3} \int x d x + \frac{1}{3} \int \frac{x}{1 + x^{2}} d x$

⇒ $I = \frac{x^{3}}{3} t a n^{- 1} x - \frac{x^{2}}{6} + \frac{1}{3} \int \frac{x}{1 + x^{2}} d x$

Now, put $1 + x^{2} = t$

⇒ $2 x d x = d t$

⇒ $x d x = \frac{d t}{2}$

⇒ $I = \frac{x^{3}}{3} t a n^{- 1} x - \frac{x^{2}}{6} + \frac{1}{6} \int \frac{d t}{t}$

⇒ $I = \frac{x^{3}}{3} t a n^{- 1} x - \frac{x^{2}}{6} + \frac{1}{6} \log | t | + C$

⇒ $I = \frac{x^{3}}{3} t a n^{- 1} x - \frac{x^{2}}{6} + \frac{1}{6} \log | 1 + x^{2} | + C$

14. Find $\int \sqrt{10 - 4 x + 4 x^{2}} d x$

Ans: Let $I = \int \sqrt{10 - 4 x + 4 x^{2}} d x$

⇒ $I = \int \sqrt{4 x^{2} - 4 x + 10} d x$

⇒ $I = \int \sqrt{4 (x^{2} - x + \frac{10}{4})} d x$

⇒ $I = 2 \int \sqrt{[x^{2} - x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + \frac{10}{4}]} d x$

⇒ $I = 2 \int \sqrt{[{(x - \frac{1}{2})}^{2} - \frac{1}{4} + \frac{10}{4}]} d x$

⇒ $I = 2 \int \sqrt{[{(x - \frac{1}{2})}^{2} + \frac{9}{4}]} d x$

⇒ $I = 2 \int \sqrt{[{(x - \frac{1}{2})}^{2} + {(\frac{3}{2})}^{2}]} d x$

⇒ $I = [\frac{(x - \frac{1}{2})}{2} \sqrt{4 x^{2} - 4 x + 10} + \frac{{(\frac{3}{2})}^{2}}{2} \log | (x - \frac{1}{2}) + \sqrt{4 x^{2} - 4 x + 10} |]$

⇒ $I = (\frac{2 x - 1}{4}) \sqrt{4 x^{2} - 4 x + 10} + \frac{9}{4} \log | (x - \frac{1}{2}) + \sqrt{4 x^{2} - 4 x + 10} |$

Long Answer (L.A.)

15. Evaluate $\int \frac{x^{2} d x}{x^{4} + x^{2} - 2}$

Ans: Let $I = \int \frac{x^{2} d x}{x^{4} + x^{2} - 2}$

Now, put $x^{2} = t$

Therefore,

⇒ $\frac{x^{2}}{x^{4} + x^{2} - 2} = \frac{t}{t^{2} + t - 2} = \frac{t}{(t + 2) (t - 1)}$

⇒ $\frac{t}{(t + 2) (t - 1)} = \frac{A}{t + 2} + \frac{B}{t - 1}$

⇒ $\frac{t}{(t + 2) (t - 1)} = \frac{A t - A + B t + 2 B}{(t + 2) (t - 1)}$

⇒ $t = (A + B) t + 2 B - A$

On comparing both side, we get

⇒ $A + B = 1$ and $2 B - A = 0$

On solving these equations, we get

⇒ $B = \frac{1}{3}$ and $A = \frac{2}{3}$

So, $\frac{x^{2}}{x^{4} + x^{2} - 2} = \frac{2}{3} \frac{1}{x^{2} + 2} + \frac{1}{3} \frac{1}{x^{2} - 1}$

Therefore,

⇒ $I = \int (\frac{2}{3} \frac{1}{x^{2} + 2} + \frac{1}{3} \frac{1}{x^{2} - 1}) d x$

⇒ $I = \frac{2}{3} \int \frac{1}{x^{2} + {(\sqrt{2})}^{2}} d x + \frac{1}{3} \int \frac{1}{x^{2} - 1} d x$

⇒ $I = \frac{2}{3} \times \frac{1}{\sqrt{2}} t a n^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{3} \times \frac{1}{2} \log | \frac{x - 1}{x + 1} | + C$

⇒ $I = \frac{\sqrt{2}}{3} t a n^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{6} \log | \frac{x - 1}{x + 1} | + C$

Hence, $\int \frac{x^{2} d x}{x^{4} + x^{2} - 2} = \frac{\sqrt{2}}{3} t a n^{- 1} \frac{x}{\sqrt{2}} + \frac{1}{6} \log | \frac{x - 1}{x + 1} | + C .$

16. Evaluate $\int \frac{x^{3} + x}{x^{4} - 9} d x$

Ans: Let $I = \int \frac{x^{3} + x}{x^{4} - 9} d x$

⇒ $I = \int \frac{x^{3}}{x^{4} - 9} d x + \int \frac{x}{x^{4} - 9} d x$

⇒ $I = I_{1} + I_{2}$ ………….. (i)

where $I_{1} = \int \frac{x^{3}}{x^{4} - 9} d x$ and $I_{2} = \int \frac{x}{x^{4} - 9} d x$

Now, $I_{1} = \int \frac{x^{3}}{x^{4} - 9} d x$

Put $x^{4} - 9 = t$

⇒ $4 x^{3} d x = d t$

⇒ $x^{3} d x = \frac{d t}{4}$

Therefore,

⇒ $I_{1} = \frac{1}{4} \int \frac{d t}{t}$

⇒ $I_{1} = \frac{1}{4} \log | t | + C_{1}$

⇒ $I_{1} = \frac{1}{4} \log | x^{4} - 9 | + C_{1}$ ……….. (ii)

Now, $I_{2} = \int \frac{x}{x^{4} - 9} d x$

Put $x^{2} = t$

⇒ $2 x d x = d t$

⇒ $x d x = \frac{d t}{2}$

Therefore,

⇒ $I_{2} = \frac{1}{2} \int \frac{d t}{t^{2} - 3^{2}}$

⇒ $I_{2} = \frac{1}{2} \times \frac{1}{2 \times 3} \log | \frac{t - 3}{t + 3} | + C_{2}$

⇒ $I_{2} = \frac{1}{12} \log | \frac{x^{2} - 3}{x^{2} + 3} | + C_{2}$

Now, put the value of $I_{1}$ and $I_{2}$ in eq (i), we get

⇒ $I = \frac{1}{4} \log | x^{4} - 9 | + C_{1} + \frac{1}{12} \log | \frac{x^{2} - 3}{x^{2} + 3} | + C_{2}$

⇒ $I = \frac{1}{4} \log | x^{4} - 9 | + \frac{1}{12} \log | \frac{x^{2} - 3}{x^{2} + 3} | + C$

17. Show that $\underset{0}{\int^{\frac{π}{2}}} \frac{s i n^{2} x d x}{\sin x + \cos x} = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

Ans: let $I = \underset{0}{\int^{\frac{π}{2}}} \frac{s i n^{2} x d x}{\sin x + \cos x}$ ………. (i)

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{s i n^{2} (\frac{π}{2} - x) d x}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)}$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{c o s^{2} x d x}{\cos x + \sin x}$ …………. (ii)

Adding eq (i) and (ii), we get

⇒ $2 I = \underset{0}{\int^{\frac{π}{2}}} \frac{(s i n^{2} x + c o s^{2} x) d x}{\cos x + \sin x}$

⇒ $2 I = \underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\cos x + \sin x}$

⇒ $2 I = \frac{1}{\sqrt{2}} \underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\cos x \frac{1}{\sqrt{2}} + \sin x . \frac{1}{\sqrt{2}}}$

⇒ $2 I = \frac{1}{\sqrt{2}} \underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\cos x \cos \frac{π}{4} + \sin x . sin \frac{π}{4}}$

⇒ $2 I = \frac{1}{\sqrt{2}} \underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\cos (x - \frac{π}{4})}$

⇒ $2 I = \frac{1}{\sqrt{2}} \underset{0}{\int^{\frac{π}{2}}} \sec (x - \frac{π}{4}) d x$

⇒ $2 I = \frac{1}{\sqrt{2}} {[\log {\sec (x - \frac{π}{4}) + \tan (x - \frac{π}{4})}]}_{0}^{\frac{π}{2}}$

⇒ $2 I = \frac{1}{\sqrt{2}} [\log {\sec (\frac{π}{2} - \frac{π}{4}) + \tan (\frac{π}{2} - \frac{π}{4})} - \log {\sec (- \frac{π}{4}) + \tan (- \frac{π}{4})}]$

⇒ $2 I = \frac{1}{\sqrt{2}} [\log {\sec (\frac{π}{4}) + \tan (\frac{π}{4})} - \log {\sec (\frac{π}{4}) - \tan (\frac{π}{4})}]$

⇒ $2 I = \frac{1}{\sqrt{2}} [\log (\sqrt{2} + 1) - \log (\sqrt{2} - 1)]$

⇒ $2 I = \frac{1}{\sqrt{2}} \times \log | \frac{\sqrt{2} + 1}{\sqrt{2} - 1} |$

⇒ $2 I = \frac{1}{\sqrt{2}} \times \log | \frac{{(\sqrt{2} + 1)}^{2}}{2 - 1} |$

⇒ $2 I = \frac{1}{\sqrt{2}} \times 2 \log (\sqrt{2} + 1)$

⇒ $I = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

Hence proved

18. Find $\underset{0}{\int^{1}} x {(t a n^{- 1} x)}^{2} d x$

Ans: Let $I = \underset{0}{\int^{1}} x {(t a n^{- 1} x)}^{2} d x$

Applying integrating by parts, we get

⇒ $I = {[\frac{x^{2}}{2} {(t a n^{- 1} x)}^{2}]}_{0}^{1} - \underset{0}{\int^{1}} \frac{2 t a n^{- 1} x}{1 + x^{2}} . \frac{x^{2}}{2} d x$

⇒ $I = [\frac{1}{2} {(t a n^{- 1} 1)}^{2} - 0] - \underset{0}{\int^{1}} \frac{t a n^{- 1} x}{1 + x^{2}} x^{2} d x$

⇒ $I = [\frac{1}{2} {(\frac{π}{4})}^{2}] - \underset{0}{\int^{1}} \frac{(x^{2} + 1 - 1) t a n^{- 1} x}{1 + x^{2}} d x$

⇒ $I = \frac{π^{2}}{32} - \underset{0}{\int^{1}} \frac{(x^{2} + 1 - 1) t a n^{- 1} x}{1 + x^{2}} d x$

⇒ $I = \frac{π^{2}}{32} - \underset{0}{\int^{1}} \frac{(x^{2} + 1) t a n^{- 1} x}{1 + x^{2}} d x + \underset{0}{\int^{1}} \frac{t a n^{- 1} x}{1 + x^{2}} d x$

⇒ $I = \frac{π^{2}}{32} - \underset{0}{\int^{1}} t a n^{- 1} x d x + \underset{0}{\int^{1}} \frac{t a n^{- 1} x}{1 + x^{2}} d x$

⇒ $I = \frac{π^{2}}{32} - I_{1} + I_{2}$ ……….. (i)

Where, $I_{1} = \underset{0}{\int^{1}} t a n^{- 1} x d x$ and $I_{2} = \underset{0}{\int^{1}} \frac{t a n^{- 1} x}{1 + x^{2}} d x$

Now, $I_{1} = \underset{0}{\int^{1}} t a n^{- 1} x d x$

⇒ $I_{1} = {[x t a n^{- 1} x]}_{0}^{1} - \underset{0}{\int^{1}} \frac{x}{1 + x^{2}} d x$

⇒ $I_{1} = [1. t a n^{- 1} 1 - 0] - \underset{0}{\int^{1}} \frac{x}{1 + x^{2}} d x$

⇒ $I_{1} = \frac{π}{4} - \underset{0}{\int^{1}} \frac{x}{1 + x^{2}} d x$

Now, put $1 + x^{2} = t$

⇒ $2 x d x = d t$

⇒ $x d x = \frac{d t}{2}$

And at $x = 0 \to t = 1$

⇒ at $x = 1 \to t = 2$

Therefore,

⇒ $I_{1} = \frac{π}{4} - \frac{1}{2} \underset{1}{\int^{2}} \frac{d t}{t}$

⇒ $I_{1} = \frac{π}{4} - \frac{1}{2} {[\log t]}_{1}^{2}$

⇒ $I_{1} = \frac{π}{4} - \frac{1}{2} [\log 2 - \log 1]$

⇒ $I_{1} = \frac{π}{4} - \frac{1}{2} \log 2$ ………….. (ii)

Now, $I_{2} = \underset{0}{\int^{1}} \frac{t a n^{- 1} x}{1 + x^{2}} d x$

Put $t a n^{- 1} x = t$

⇒ $\frac{1}{1 + x^{2}} d x = d t$

And at $x = 0 \to t = 0$

⇒ at $x = 1 \to t = \frac{π}{4}$

Therefore,

⇒ $I_{2} = \underset{0}{\int^{\frac{π}{4}}} t d t$

⇒ $I_{2} = {[\frac{t^{2}}{2}]}_{0}^{\frac{π}{4}}$

⇒ $I_{2} = \frac{1}{2} {[t^{2}]}_{0}^{\frac{π}{4}}$

⇒ $I_{2} = \frac{1}{2} [\frac{π^{2}}{16} - 0]$

⇒ $I_{2} = \frac{π^{2}}{32}$

Now, put the value of $I_{2}$ and $I_{2}$ in eq(i), we get

⇒ $I = \frac{π^{2}}{32} - \frac{π}{4} + \frac{1}{2} \log 2 + \frac{π^{2}}{32}$

⇒ $I = \frac{π^{2}}{16} - \frac{π}{4} + \frac{1}{2} \log 2$

Hence, $\underset{0}{\int^{1}} x {(t a n^{- 1} x)}^{2} d x = \frac{π^{2}}{16} - \frac{π}{4} + \frac{1}{2} \log 2.$

19. Evaluate $\underset{- 1}{\int^{2}} f (x) d x$ , where $f (x) = | x + 1 | + | x | + | x - 1 |$ .

Ans: here, we can redefine $f (x)$ as

f\left( x \right)=\left\{ $\begin{matrix} 2 - x, i f - 1 < x \leq 0 \\ x + 2, i f 0 < x \leq 1 \\ 3 x, i f 1 < x \leq 2 \end{matrix}$ \right.$

Therefore, $\underset{- 1}{\int^{2}} f (x) d x = \underset{- 1}{\int^{0}} (2 - x) d x + \underset{0}{\int^{1}} (x + 2) d x + \underset{1}{\int^{2}} 3 x d x$

⇒ $\underset{- 1}{\int^{2}} f (x) d x = {[2 x - \frac{x^{2}}{2}]}_{- 1}^{0} + {[\frac{x^{2}}{2} + 2 x]}_{0}^{1} + 3 {[\frac{x^{2}}{2}]}_{1}^{2}$

⇒ $\underset{- 1}{\int^{2}} f (x) d x = [0 - (- 2 - \frac{1}{2})] + [(\frac{1}{2} + 2) - 0] + 3 [2 - \frac{1}{2}]$

⇒ $\underset{- 1}{\int^{2}} f (x) d x = \frac{5}{2} + \frac{5}{2} + \frac{9}{2}$

⇒ $\underset{- 1}{\int^{2}} f (x) d x = \frac{19}{2}$ .

Objective Type Questions

Choose the correct answer from the given four option in each of the Examples from 20 to 30.

20. $\int e^{x} (\cos x - \sin x) d x$ is equal to

(A) $e^{x} \cos x + C$

(B) $e^{x} \sin x + C$

(D) $- e^{x} \sin x + C$

Ans: here, as we know that $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$

Now, if $f (x) = \cos x$ , then $f^{'} (x) = - \sin x$

Therefore, $\int e^{x} (\cos x - \sin x) d x = e^{x} \cos x + C$

Hence, option (A) is the correct answer.

21. $\int \frac{d x}{s i n^{2} x c o s^{2} x}$ is equal to

(A) $\tan x + \cot x + C$

(B) ${(\tan x + \cot x)}^{2} + C$

(D) ${(\tan x - \cot x)}^{2} + C$

Ans: Let $I = \int \frac{d x}{s i n^{2} x c o s^{2} x}$

⇒ $I = \int \frac{(s i n^{2} x + c o s^{2} x) d x}{s i n^{2} x c o s^{2} x}$

⇒ $I = \int \frac{s i n^{2} x d x}{s i n^{2} x c o s^{2} x} + \int \frac{c o s^{2} x d x}{s i n^{2} x c o s^{2} x}$

⇒ $I = \int \frac{d x}{c o s^{2} x} + \int \frac{d x}{s i n^{2} x}$

⇒ $I = \int s e c^{2} x d x + \int c o s e c^{2} x d x$

⇒ $I = \tan x - \cot x + C$

Hence, option (C) is the correct answer.

22. If $\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = a x + b \log | 4 e^{x} + 5 e^{- x} | + C$ , then

(A) $a = - \frac{1}{8}, b = \frac{7}{8}$

(B) $a = \frac{1}{8}, b = \frac{7}{8}$

(D) $a = \frac{1}{8}, b = - \frac{7}{8}$

Ans: here, we have $\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x = a x + b \log | 4 e^{x} + 5 e^{- x} | + C$

Now, differentiating both sides, we have

⇒ $\frac{d}{d x} (\int \frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} d x) = \frac{d}{d x} (a x + b \log | 4 e^{x} + 5 e^{- x} | + C)$

⇒ $\frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} = a + b \times \frac{1}{4 e^{x} + 5 e^{- x}} \times (4 e^{x} - 5 e^{- x})$

⇒ $\frac{3 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}} = a + b (\frac{4 e^{x} - 5 e^{- x}}{4 e^{x} + 5 e^{- x}})$

⇒ $(3 e^{x} - 5 e^{- x}) = a (4 e^{x} + 5 e^{- x}) + b (4 e^{x} - 5 e^{- x})$

⇒ $(3 e^{x} - 5 e^{- x}) = 4 a e^{x} + 5 a e^{- x} + 4 b e^{x} - 5 b e^{- x}$

⇒ $3 e^{x} - 5 e^{- x} = 4 e^{x} (a + b) - 5 e^{- x} (- a + b)$

Now, comparing both sides, we get

⇒ $4 (a + b) = 3$ and $(- a + b) = 1$

⇒ $(a + b) = \frac{3}{4}$ …… (i) and $(- a + b) = 1$ ………(ii)

Adding eq (i) and (ii), we get

⇒ $2 b = \frac{7}{4}$

⇒ $b = \frac{7}{8}$ .

Now, put the value of $b$ in eq (ii),

⇒ $- a + \frac{7}{8} = 1$

⇒ $- a = 1 - \frac{7}{8}$

⇒ $a = - \frac{1}{8}$

Here, $a = - \frac{1}{8}$ and $b = \frac{7}{8}$

Hence, option (A) is the correct answer.

23: $\underset{a + c}{\int^{b + c}} f (x) d x$ is equal to

(A) $\underset{a}{\int^{b}} f (x - c) d x$

(B) $\underset{a}{\int^{b}} f (x + c) d x$

(D) $\underset{a - c}{\int^{b - c}} f (x) d x$

Ans: Let $I = \underset{a + c}{\int^{b + c}} f (x) d x$

Now, put $x = t + c$

⇒ $d x = d t$

And at $x = a + c$ ⇒ $t = a$

⇒ at $x = b + c$ ⇒ $t = b$

Therefore,

⇒ $I = \underset{a}{\int^{b}} f (t + c) d t$

⇒ $I = \underset{a}{\int^{b}} f (x + c) d x$

Hence, option (B) is the correct answer.

24. If $f$ and $g$ are continuous function in $[0, 1]$ satisfying $f (x) = f (a - x)$ and $g (x) + g (a - x) = a,$ then $\underset{0}{\int^{a}} f (x) . g (x) d x$ is equal to

(A) $\frac{a}{2}$

(B) $\frac{a}{2} \underset{0}{\int^{a}} f (x) d x$

(D) $a \underset{0}{\int^{a}} f (x) d x$

Ans: Let $I = \underset{0}{\int^{a}} f (x) . g (x) d x$

⇒ $I = \underset{0}{\int^{a}} f (x - a) . g (x - a) d x$

⇒ $I = \underset{0}{\int^{a}} f (x) . (a - g (x)) d x$

⇒ $I = \underset{0}{\int^{a}} [a f (x) - f (x) . g (x)] d x$

⇒ $I = a \underset{0}{\int^{a}} f (x) d x - \underset{0}{\int^{a}} f (x) . g (x) d x$

⇒ $I = a \underset{0}{\int^{a}} f (x) d x - I$

⇒ $2 I = a \underset{0}{\int^{a}} f (x) d x$

⇒ $I = \frac{a}{2} \underset{0}{\int^{a}} f (x) d x$

Hence, option (B) is the correct answer.

25: If $x = \underset{0}{\int^{y}} \frac{d t}{\sqrt{1 + 9 t^{2}}}$ and $\frac{d^{2} y}{d x^{2}} = a y$ , then $a$ is equal to

(A) 3

(B) 6

(D) 1

Ans: here, we have $x = \underset{0}{\int^{y}} \frac{d t}{\sqrt{1 + 9 t^{2}}}$

Differentiate w. r.t $y$ , we get

⇒ $\frac{d x}{d y} = \frac{1}{\sqrt{1 + 9 y^{2}}}$

⇒ $\frac{d y}{d x} = \sqrt{1 + 9 y^{2}}$

Now, differentiate w. r.t $x$ , we get

⇒ $\frac{d^{2} y}{d x^{2}} = \frac{18 y}{2 \sqrt{1 + 9 y^{2}}} . \frac{d y}{d x}$

⇒ $\frac{d^{2} y}{d x^{2}} = \frac{9 y}{\sqrt{1 + 9 y^{2}}} . \sqrt{1 + 9 y^{2}}$

⇒ $\frac{d^{2} y}{d x^{2}} = 9 y$

Thus, value of $x$ is 9.

Hence, option (C) is the correct answer.

26. $\underset{- 1}{\int^{1}} \frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1} d x$ is equal to

(A) $\log 2$

(B) $2 \log 2$

(D) $4 \log 2$

Ans: Let $I = \underset{- 1}{\int^{1}} \frac{x^{3} + | x | + 1}{x^{2} + 2 | x | + 1} d x$

⇒ $I = \underset{- 1}{\int^{1}} \frac{x^{3}}{x^{2} + 2 | x | + 1} d x + \underset{- 1}{\int^{1}} \frac{| x | + 1}{x^{2} + 2 | x | + 1} d x$

Here, $\underset{- 1}{\int^{1}} \frac{x^{3}}{x^{2} + 2 | x | + 1} d x$ is odd function and $\underset{- 1}{\int^{1}} \frac{| x | + 1}{x^{2} + 2 | x | + 1} d x$ is even function.

⇒ $I = 0 + 2 \underset{0}{\int^{1}} \frac{| x | + 1}{{(| x | + 1)}^{2}} d x$

⇒ $I = 2 \underset{0}{\int^{1}} \frac{x + 1}{{(x + 1)}^{2}} d x$

⇒ $I = 2 \underset{0}{\int^{1}} \frac{1}{(x + 1)} d x$

⇒ $I = 2 {[\log | x + 1 |]}_{0}^{1}$

⇒ $I = 2 [\log 2 - \log 1]$

⇒ $I = 2 \log 2$

Hence, option (B) is the correct answer.

27. If $\underset{0}{\int^{1}} \frac{e^{t}}{1 + t} d t = a$ , then $\underset{0}{\int^{1}} \frac{e^{t}}{{(1 + t)}^{2}} d t$ is equal to

(A) $a - 1 + \frac{e}{2}$

(B) $a + 1 - \frac{e}{2}$

(D) $a + 1 + \frac{e}{2}$

Ans: here, we have $\underset{0}{\int^{1}} \frac{e^{t}}{1 + t} d t = a$ ,

⇒ $a = \underset{0}{\int^{1}} \frac{e^{t}}{1 + t} d t$

Applying integrating by parts, we get

⇒ $a = {[e^{t} \times \frac{1}{(1 + t)}]}_{0}^{1} + \underset{0}{\int^{1}} \frac{e^{t}}{{(1 + t)}^{2}} d t$

⇒ $a = [\frac{e}{2} - 1] + \underset{0}{\int^{1}} \frac{e^{t}}{{(1 + t)}^{2}} d t$

⇒ $\underset{0}{\int^{1}} \frac{e^{t}}{{(1 + t)}^{2}} d t = a + 1 - \frac{e}{2}$

Hence, option (B) is the correct answer.

28. $\underset{- 2}{\int^{2}} | x \cos π x | d x$ is equal to

(A) $\frac{8}{π}$

(B) $\frac{4}{π}$ .

(D) $\frac{1}{π}$

Ans: Let $I = \underset{- 2}{\int^{2}} | x \cos π x | d x$

Here, it is a even function, therefore

⇒ $I = 2 \underset{0}{\int^{2}} | x \cos π x | d x$

⇒ $\frac{I}{2} = \underset{0}{\int^{\frac{1}{2}}} x \cos π x d x + \underset{\frac{1}{2}}{\int^{\frac{3}{2}}} - x \cos π x d x + \underset{\frac{3}{2}}{\int^{2}} x \cos π x d x$

⇒ $\frac{I}{2} = {[\frac{x}{π} \sin π x + \frac{1}{π^{2}} \cos π x]}_{0}^{\frac{1}{2}} - {[\frac{x}{π} \sin π x + \frac{1}{π^{2}} \cos π x]}_{\frac{1}{2}}^{\frac{3}{2}} + {[\frac{x}{π} \sin π x + \frac{1}{π^{2}} \cos π x]}_{\frac{3}{2}}^{0}$

⇒ $\frac{I}{2} = [\frac{1}{2 π} - \frac{1}{π^{2}}] - [- \frac{3}{2 π} - \frac{1}{2 π}] + [\frac{1}{π^{2}} + \frac{3}{2 π}]$

⇒ $\frac{I}{2} = \frac{1}{2 π} - \frac{1}{π^{2}} + \frac{3}{2 π} + \frac{1}{2 π} + \frac{1}{π^{2}} + \frac{3}{2 π}$

⇒ $\frac{I}{2} = \frac{8}{2 π}$

⇒ $I = \frac{8}{π}$

Hence, option (A) is the correct answer.

Fill in the blanks in each of the 29 to 32.

29. $\int \frac{s i n^{6} x}{c o s^{8} x} d x =$ …………..

Ans: Let $I = \int \frac{s i n^{6} x}{c o s^{8} x} d x$

⇒ $I = \int \frac{s i n^{6} x}{c o s^{6} x . c o s^{2} x} d x$

⇒ $I = \int t a n^{6} x . s e c^{2} x d x$

Put $\tan x = t$

⇒ $s e c^{2} x d x = d t$

Therefore,

⇒ $I = \int t^{6} d t$

⇒ $I = \frac{t^{7}}{7} + C$

⇒ $I = \frac{t a n^{7} x}{7} + C$

Hence, $\int \frac{s i n^{6} x}{c o s^{8} x} d x = \frac{t a n^{7} x}{7} + C$ .

30. $\underset{- a}{\int^{a}} f (x) d x = 0$ if $f$ is an ……………. function.

Ans: As we know that by the property of integration, $\underset{- a}{\int^{a}} f (x) d x = 0$ if $f$ is an odd function.

31. $\underset{0}{\int^{2 a}} f (x) d x = 2 \underset{0}{\int^{a}} f (x) d x,$ if $f (2 a - x) =$ ………..

Ans: As, we know that by the property of integration, we have

⇒ $\underset{0}{\int^{2 a}} f (x) d x = 2 \underset{0}{\int^{a}} f (x) d x,$ if $f (2 a - x) = f (x)$

$= 0,$ if $f (2 a - x) = - f (x)$

Hence, $\underset{0}{\int^{2 a}} f (x) d x = 2 \underset{0}{\int^{a}} f (x) d x,$ if $f (2 a - x) = f (x) .$

32. $\underset{0}{\int^{\frac{π}{2}}} \frac{s i n^{n} x}{s i n^{n} x + c o s^{n} x} d x =$ …………

Ans: Let $I = \underset{0}{\int^{\frac{π}{2}}} \frac{s i n^{n} x}{s i n^{n} x + c o s^{n} x} d x$ …………….. (i)

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{s i n^{n} (\frac{π}{2} - x)}{s i n^{n} (\frac{π}{2} - x) + c o s^{n} (\frac{π}{2} - x)} d x$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{c o s^{n} x}{c o s^{n} x + s i n^{n} x} d x$ ……………… (ii)

Adding eq (i) and (ii), we get

⇒ $2 I = \underset{0}{\int^{\frac{π}{2}}} d x$

⇒ $2 I = {[x]}_{0}^{\frac{π}{2}}$

⇒ $2 I = \frac{π}{2}$

⇒ $I = \frac{π}{4}$

Hence, $\underset{0}{\int^{\frac{π}{2}}} \frac{s i n^{n} x}{s i n^{n} x + c o s^{n} x} d x = \frac{π}{4} .$

Exercise 7.3

Short Answer (S.A)

Verify the following:

1. $\int \frac{2 x - 1}{2 x + 3} d x = x - \log | {(2 x - 3)}^{2} | + C$

Ans: Let $I = \int \frac{2 x - 1}{2 x + 3} d x$

$\Rightarrow I = \int \frac{2 x + 3 - 3 - 1}{2 x + 3} d x$

$\Rightarrow I = \int \frac{2 x + 3 - 4}{2 x + 3} d x$

$\Rightarrow I = \int (1 - \frac{4}{2 x + 3}) d x$

$\Rightarrow I = \int d x - \int \frac{4}{2 x + 3} d x$

$\Rightarrow I = x - \int \frac{4}{2 (x + \frac{3}{2})} d x$

$\Rightarrow I = x - 2 \log (x + \frac{3}{2}) + C$

$\Rightarrow I = x - 2 log (\frac{2 x + 3}{2}) + C$

$\Rightarrow I = x - 2 [\log (2 x + 3) - \log 2] + C$

$\Rightarrow I = x - 2 log (2 x + 3) + 2 \log 2 + C$

$\Rightarrow I = x - log | {(2 x + 3)}^{2} | + C$

Thus, $\int \frac{2 x - 1}{2 x + 3} d x = x - \log | {(2 x - 3)}^{2} | + C$

Hence proved.

2. $\int \frac{2 x + 3}{x^{2} + 3 x} d x = \log | x^{2} + 3 x | + C$

Ans: let $I = \int \frac{2 x + 3}{x^{2} + 3 x} d x$

Put $x^{2} + 3 x = t$

⇒ $(2 x + 3) d x = d t$

$∴ I = \int \frac{1}{t} d t$

⇒ $I = \log | t | + C$

⇒ $I = \log | (x^{2} + 3 x) | + C$

Thus, $\int \frac{2 x + 3}{x^{2} + 3 x} d x = log | x^{2} + 3 x | + C$

Hence proved.

Evaluate the Following:

3. $\int \frac{(x^{2} + 2) d x}{x + 1}$

Ans: Let $I = \int \frac{(x^{2} + 2) d x}{x + 1}$

⇒ $I = \int (x - 1 + \frac{3}{x + 1}) d x$

⇒ $I = \int x d x - \int d x + \int \frac{3}{x + 1} d x$

⇒ $I = \frac{x^{2}}{2} - x + 3 log | x + 1 | + C$

Hence, $\int \frac{(x^{2} + 2) d x}{x + 1} = \frac{x^{2}}{2} - x + 3 log | x + 1 | + C .$

4. $\int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} d x$

Ans: Let $I = \int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} d x$

⇒ $I = \int \frac{e^{\log x^{6}} - e^{\log x^{5}}}{e^{\log x^{4}} - e^{\log x^{3}}} d x$

⇒ $I = \int \frac{x^{6} - x^{5}}{x^{4} - x^{3}} d x$ [ $e^{\log x} = x]$

⇒ $I = \int \frac{x^{5} (x - 1)}{x^{3} (x - 1)} d x$

⇒ $I = \int x^{2} d x$

⇒ $I = \frac{x^{3}}{3} + C$

Hence, $\int \frac{e^{6 \log x} - e^{5 \log x}}{e^{4 \log x} - e^{3 \log x}} d x = \frac{x^{3}}{3} + C .$

5. $\int \frac{(1 + \cos x)}{x + \sin x} d x$

Ans: Let $I = \int \frac{(1 + \cos x)}{x + \sin x} d x$

Put $x + \sin x = t$

⇒ $(1 + \cos x) d x = d t$

⇒ $I = \int \frac{1}{t} d x$

⇒ $I = \log | t | + C$

⇒ $I = \log | (x + \cos x) | + C$

6. $\int \frac{d x}{1 + \cos x}$

Ans: Let $I = \int \frac{d x}{1 + \cos x}$

⇒ $I = \int \frac{d x}{2 c o s^{2} \frac{x}{2}}$

⇒ $I = \frac{1}{2} \int \frac{d x}{c o s^{2} \frac{x}{2}}$

⇒ $I = \frac{1}{2} \int s e c^{2} \frac{x}{2} d x$

⇒ $I = \frac{1}{2} \times 2 \tan \frac{x}{2} + C$ [ $\int s e c^{2} x d x = \tan x + C]$

⇒ $I = \tan \frac{x}{2} + C$

Hence, $\int \frac{d x}{1 + \cos x} = \tan \frac{x}{2} + C$ .

7. $\int t a n^{2} x . s e c^{4} x d x$

Ans: Let $I = \int t a n^{2} x . s e c^{4} x d x$

⇒ $I = \int t a n^{2} x . s e c^{2} x . s e c^{2} x d x$

⇒ $I = \int t a n^{2} x (1 + t a n^{2} x) s e c^{2} x d x$

Put $tan x = t$

⇒ $se c^{2} x dx = dt$

Therefore,

⇒ $I = \int t^{2} (1 + t^{2}) d t$

⇒ $I = \int (t^{2} + t^{4}) d t$

⇒ $I = \frac{t^{3}}{3} + \frac{t^{5}}{5} + C$

⇒ $I = \frac{t a n^{3} x}{3} + \frac{t a n^{5} x}{5} + C$

8. $\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$

Ans: Let $I = \int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x$

⇒ $I = \int \frac{\sin x + \cos x}{\sqrt{s i n^{2} x + c o s^{2} x + 2. \sin x \cos x}} d x$

⇒ $I = \int \frac{\sin x + \cos x}{\sqrt{{(\sin x + \cos x)}^{2}}} d x$

⇒ $I = \int \frac{(\sin x + \cos x)}{(\sin x + \cos x)} d x$

⇒ $I = \int d x$

⇒ $I = x + C$

Hence, $\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2 x}} d x = x + C$ .

9. $\int \sqrt{1 + \sin x} d x$

Ans: Let $I = \int \sqrt{1 + \sin x} d x$

⇒ $I = \int \sqrt{s i n^{2} \frac{x}{2} + c o s^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}} d x$

⇒ $I = \int \sqrt{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}} d x$

⇒ $I = \int (\sin \frac{x}{2} + \cos \frac{x}{2}) d x$

⇒ $I = \int \sin \frac{x}{2} d x + \int \cos \frac{x}{2} d x$

⇒ $I = - 2 cos \frac{x}{2} + 2 sin \frac{x}{2} + C$

Hence, $\int \sqrt{1 + \sin x} d x = - 2 cos \frac{x}{2} + 2 sin \frac{x}{2} + C .$

10. $\int \frac{x}{\sqrt{x} + 1} d x$

Ans: Let $I = \int \frac{x}{\sqrt{x} + 1} dx$

Put $\sqrt{x} = z$

⇒ $\frac{1}{2 \sqrt{x}} d x = d z$

⇒ $d x = 2 \sqrt{x} d z$

⇒ $d x = 2 z d z$

Therefore,

⇒ $I = \int \frac{z^{2}}{z + 1} \times 2 z dz$

⇒ $I = 2 \int \frac{z^{3}}{z + 1} dz$

⇒ $I = 2 \int \frac{z^{3} + 1 - 1}{z + 1} dz$

⇒ $I = 2 \int \frac{z^{3} + 1}{z + 1} dz - 2 \int \frac{1}{z + 1} dz$

⇒ $I = 2 \int \frac{(z + 1) (z^{2} - z + 1)}{z + 1} dz - 2 \int \frac{1}{z + 1} dz$

⇒ $I = 2 \int (z^{2} - z + 1) dz - 2 \int \frac{1}{z + 1} dz$

⇒ $I = 2 (\frac{z^{3}}{3} - \frac{z^{2}}{2} + z) - 2 log | z + 1 | + C$

⇒ $I = 2 (\frac{x \sqrt{x}}{3} - \frac{x}{2} + \sqrt{x}) - 2 log | \sqrt{x} + 1 | + C$

⇒ $I = 2 (\frac{x \sqrt{x}}{3} - \frac{x}{2} + \sqrt{x} - log | \sqrt{x} + 1 |) + C$

Hence, $\int \frac{x}{\sqrt{x} + 1} dx = 2 (\frac{x \sqrt{x}}{3} - \frac{x}{2} + \sqrt{x} - log | \sqrt{x} + 1 |) + C .$

11. $\int \sqrt{\frac{a + x}{a - x}} d x$

Ans: Let $I = \int \sqrt{\frac{a + x}{a - x}} d x$

⇒ $I = \int \sqrt{\frac{(a + x) (a + x)}{(a - x) (a + x)}} d x$

⇒ $I = \int \sqrt{\frac{(a + x)^{2}}{(a^{2} - x^{2})}} d x$

⇒ $I = \int \frac{a + x}{\sqrt{(a^{2} - x^{2})}} d x$

⇒ $I = \int \frac{a}{\sqrt{(a^{2} - x^{2})}} d x + \int \frac{x}{\sqrt{(a^{2} - x^{2})}} d x$

⇒ $I = a s i n^{- 1} (\frac{x}{a}) + \int \frac{x}{\sqrt{(a^{2} - x^{2})}} d x$

Put $a^{2} - x^{2} = t^{2}$

⇒ $- 2 x d x = 2 t d t$

⇒ $x d x = - t d t$

Therefore,

⇒ $I = a s i n^{- 1} (\frac{x}{a}) + \int \frac{- t d t}{\sqrt{t^{2}}}$

⇒ $I = a s i n^{- 1} (\frac{x}{a}) - \int \frac{t d t}{t}$

⇒ $I = a s i n^{- 1} (\frac{x}{a}) - \int d t$

⇒ $I = a s i n^{- 1} (\frac{x}{a}) - t + C$

⇒ $I = a s i n^{- 1} (\frac{x}{a}) - \sqrt{(a^{2} - x^{2})} + C$

12. $\int \frac{x^{\frac{1}{2}}}{1 + x^{\frac{3}{4}}} d x$

Ans: Let $I = \int \frac{x^{\frac{1}{2}}}{1 + x^{\frac{3}{4}}} d x$

Put $x = z^{4}$

⇒ $d x = 4 z^{3} d z$

Therefore,

⇒ $I = \int \frac{{(z^{4})}^{\frac{1}{2}}}{1 + {(z^{4})}^{\frac{3}{4}}} \times 4 z^{3} d z$

⇒ $I = 4 \int \frac{z^{2} . z^{3}}{1 + z^{3}} d z$

Now, put $1 + z^{3} = t$

⇒ $3 z^{2} d z = d t$

⇒ $z^{2} d z = \frac{d t}{3}$

Therefore, we get

⇒ $I = 4 \int \frac{(t - 1)}{t} . \frac{d t}{3}$

⇒ $I = \frac{4}{3} \int (1 - \frac{1}{t}) d t$

⇒ $I = \frac{4}{3} t - \frac{4}{3} log | t | + C_{1}$

⇒ $I = \frac{4}{3} (1 + z^{3}) - \frac{4}{3} log | 1 + z^{3} | + C_{1}$

⇒ $I = \frac{4}{3} (1 + x^{\frac{3}{4}}) - \frac{4}{3} log | 1 + x^{\frac{3}{4}} | + C_{1}$

⇒ $I = \frac{4}{3} + \frac{4}{3} x^{\frac{3}{4}} - \frac{4}{3} log | 1 + x^{\frac{3}{4}} | + C_{1}$

⇒ $I = \frac{4}{3} x^{\frac{3}{4}} - \frac{4}{3} log | 1 + x^{\frac{3}{4}} | + C$

13. $\int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x$

Ans: Let $I = \int \frac{\sqrt{1 + x^{2}}}{x^{4}} d x$

⇒ $I = \int \frac{\sqrt{1 + x^{2}}}{x . x^{3}} d x$

⇒ $I = \int \sqrt{\frac{1 + x^{2}}{x^{2}}} \frac{d x}{x^{3}}$

⇒ $I = \int \sqrt{(\frac{1}{x^{2}} + 1)} \frac{d x}{x^{3}}$

Put $(\frac{1}{x^{2}} + 1) = t^{2}$

⇒ $- \frac{2}{x^{3}} d x = 2 t d t$

⇒ $\frac{d x}{x^{3}} = - t d t$

Therefore,

⇒ $I = \int \sqrt{t^{2}} (- t) d t$

⇒ $I = - \int t . t d t$

⇒ $I = - \int t^{2} d t$

⇒ $I = - \frac{t^{3}}{3} + C$

⇒ $I = - \frac{1}{3} {(\frac{1}{x^{2}} + 1)}^{\frac{3}{2}} + C$

14. $\int \frac{d x}{\sqrt{16 - 9 x^{2}}}$

Ans: Let $I = \int \frac{d x}{\sqrt{16 - 9 x^{2}}}$

⇒ $I = \int \frac{d x}{\sqrt{9 (\frac{16}{9} - x^{2})}}$

⇒ $I = \frac{1}{3} \int \frac{d x}{\sqrt{{(\frac{4}{3})}^{2} - x^{2}}}$

⇒ $I = \frac{1}{3} s i n^{- 1} (\frac{x}{\frac{4}{3}}) + C$

⇒ $I = \frac{1}{3} s i n^{- 1} (\frac{3 x}{4}) + C$

15. $\int \frac{d t}{\sqrt{3 t - 2 t^{2}}}$

Ans: Let $I = \int \frac{d t}{\sqrt{3 t - 2 t^{2}}}$

⇒ $I = \int \frac{d t}{\sqrt{- 2 (t^{2} - \frac{3 t}{2})}}$

⇒ $I = \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{- (t^{2} - \frac{3 t}{2})}}$

⇒ $I = \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{- [t^{2} - \frac{3 t}{2} + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{2}]}}$

⇒ $I = \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{- [{(t - \frac{3}{4})}^{2} - {(\frac{3}{4})}^{2}]}}$

⇒ $I = \frac{1}{\sqrt{2}} \int \frac{d t}{\sqrt{{(\frac{3}{4})}^{2} - {(t - \frac{3}{4})}^{2}}}$

⇒ $I = \frac{1}{\sqrt{2}} s i n^{- 1} (\frac{t - \frac{3}{4}}{\frac{3}{4}}) + C$

⇒ $I = \frac{1}{\sqrt{2}} s i n^{- 1} (\frac{4 t - 3}{3}) + C$

16. $\int \frac{3 x - 1}{\sqrt{x^{2} + 9}} dx$

Ans: Let $I = \int \frac{3 x - 1}{\sqrt{x^{2} + 9}} dx$

⇒ $I = \int \frac{3 x}{\sqrt{x^{2} + 9}} dx - \int \frac{dx}{\sqrt{x^{2} + 9}}$

⇒ $I = I_{1} - I_{2}$ ………………. (i),

Where, $I_{1} = \int \frac{3 x}{\sqrt{x^{2} + 9}} dx$ and $I_{2} = \int \frac{dx}{\sqrt{x^{2} + 9}}$

Therefore,

⇒ $I_{1} = \int \frac{3 x}{\sqrt{x^{2} + 9}} dx$

Put $x^{2} + 9 = t^{2}$

⇒ $2 x d x = 2 t d t$

⇒ $x d x = t d t$

⇒ $I_{1} = \int \frac{3 t dt}{\sqrt{t^{2}}}$

⇒ $I_{1} = \int \frac{3 t dt}{t}$

⇒ $I_{1} = \int 3 d t$

⇒ $I_{1} = 3 t + C_{1}$

⇒ $I_{1} = 3 \sqrt{x^{2} + 9} + C_{1}$ ……….. (ii)

Also we have,

$\Rightarrow I_{2} = \int \frac{dx}{\sqrt{x^{2} + 9}}$

$\Rightarrow I_{2} = \int \frac{dx}{\sqrt{x^{2} + 3^{2}}}$

$\Rightarrow I_{2} = log | x + \sqrt{x^{2} + 9} | + C_{2}$ ………….. (iii)

Put the value of $I_{1}$ and $I_{2}$ in eq (i), we get

$\Rightarrow I = 3 \sqrt{x^{2} + 9} + C_{1} - log | x + \sqrt{x^{2} + 9} | - C_{2}$

$\Rightarrow I = 3 \sqrt{x^{2} + 9} - log | x + \sqrt{x^{2} + 9} | + C$

17. $\int \sqrt{5 - 2 x + x^{2}} d x$

Ans: Let $I = \int \sqrt{5 - 2 x + x^{2}} d x$

⇒ $I = \int \sqrt{x^{2} - 2 x + 5} d x$ .

⇒ $I = \int \sqrt{x^{2} - 2 x + 1 + 4} d x$

⇒ $I = \int \sqrt{(x^{2} - 2 x + 1) + 2^{2}} d x$

⇒ $I = \int \sqrt{{(x - 1)}^{2} + 2^{2}} d x$

⇒ $I = \frac{x - 1}{2} \sqrt{{(x - 1)}^{2} + 2^{2}} + \frac{2^{2}}{2} log | x - 1 + \sqrt{{(x - 1)}^{2} + 2^{2}} | + C$

⇒ $I = \frac{(x - 1)}{2} \sqrt{5 - 2 x + x^{2}} + 2 log | x - 1 + \sqrt{5 - 2 x + x^{2}} | + C$ .

18. $\int \frac{x}{x^{4} - 1} d x$ .

Ans: Let $I = \int \frac{x}{x^{4} - 1} d x$

⇒ $I = \int \frac{x}{{(x^{2})}^{2} - 1} d x$

Put $x^{2} = t$ .

⇒ $2 x d x = d t$ .

⇒ $x d x = \frac{d t}{2}$

Therefore,

⇒ $I = \frac{1}{2} \int \frac{d t}{{(t)}^{2} - 1}$

⇒ $I = \frac{1}{2} . \frac{1}{2 \times 1} log | \frac{t - 1}{t + 1} | + C$

⇒ $I = \frac{1}{4} log | \frac{x^{2} - 1}{x^{2} + 1} | + C$

19. $\int \frac{x^{2}}{1 - x^{4}} d x$

Ans: Let $I = \int \frac{x^{2}}{1 - x^{4}} d x$

⇒ $I = \int \frac{\frac{x^{2}}{2} + \frac{x^{2}}{2}}{1^{2} - {(x^{2})}^{2}} d x$

⇒ $I = \int \frac{\frac{1}{2} + \frac{x^{2}}{2} - \frac{1}{2} + \frac{x^{2}}{2}}{(1 - x^{2}) (1 + x^{2})} d x$

⇒ $I = \int \frac{\frac{1}{2} (1 + x^{2}) - \frac{1}{2} (1 - x^{2})}{(1 - x^{2}) (1 + x^{2})} d x$

⇒ $I = \frac{1}{2} \int \frac{(1 + x^{2})}{(1 - x^{2}) (1 + x^{2})} d x - \frac{1}{2} \int \frac{(1 - x^{2})}{(1 - x^{2}) (1 + x^{2})} d x$

⇒ $I = \frac{1}{2} \int \frac{d x}{(1 - x^{2})} - \frac{1}{2} \int \frac{d x}{(1 + x^{2})}$

⇒ $I = \frac{1}{2} . \frac{1}{2 \times 1} log | \frac{1 + x}{1 - x} | - \frac{1}{2} t a n^{- 1} x + C$

⇒ $I = \frac{1}{4} log | \frac{1 + x}{1 - x} | - \frac{1}{2} t a n^{- 1} x + C$ .

20. $\int \sqrt{2 a x - x^{2}} d x$

Ans: Let $I = \int \sqrt{2 a x - x^{2}} d x$

⇒ $I = \int \sqrt{- (x^{2} - 2 a x)} d x$

⇒ $I = \int \sqrt{- (x^{2} - 2 a x + a^{2} - a^{2})} d x$ .

⇒ $I = \int \sqrt{- [(x^{2} - 2 a x + a^{2}) - a^{2}]} d x$ .

⇒ $I = \int \sqrt{- [{(x - a)}^{2} - a^{2}]} d x$ .

⇒ $I = \int \sqrt{a^{2} - {(x - a)}^{2}} d x$ .

⇒ $I = \frac{(x - a)}{2} \sqrt{a^{2} - {(x - a)}^{2}} + \frac{a^{2}}{2} s i n^{- 1} (\frac{x - a}{a}) + C$ .

⇒ $I = \frac{(x - a)}{2} \sqrt{2 a x - x^{2}} + \frac{a^{2}}{2} s i n^{- 1} (\frac{x - a}{a}) + C$

21. $\int \frac{s i n^{- 1} x}{{(1 - x^{2})}^{\frac{3}{2}}} d x$ .

Let $I = \int \frac{s i n^{- 1} x}{{(1 - x^{2})}^{\frac{3}{2}}} d x$ .

⇒ $I = \int \frac{s i n^{- 1} x}{(1 - x^{2}) \sqrt{1 - x^{2}}} d x$ .

Put $s i n^{- 1} x = t$ .

⇒ $\frac{1}{\sqrt{1 - x^{2}}} d x = d t$

And $s i n^{- 1} x = t$ ⇒ $x = s i n t$

$cost = \sqrt{1 - x^{2}}$

Therefore,

$I = \int \frac{t . d t}{(1 - s i n^{2} t)}$

⇒ $I = \int \frac{t . d t}{c o s^{2} t}$

⇒ $I = \int t . s e c^{2} t d t$

⇒ $I = t \int s e c^{2} t d t - \int (\frac{d}{d t} t . \int s e c^{2} t d t) d t$

⇒ $I = t tant - \int tantdt$

⇒ $I = t tant + log | cost | + C$

⇒ $I = s i n^{- 1} x \times \frac{\sin t}{\cos t} + log | \sqrt{1 - x^{2}} | + C$

⇒ $I = \frac{x s i n^{- 1} x}{\sqrt{1 - x^{2}}} + log | \sqrt{1 - x^{2}} | + C$

Hence, $\int \frac{s i n^{- 1} x}{{(1 - x^{2})}^{\frac{3}{2}}} d x = \frac{x s i n^{- 1} x}{\sqrt{1 - x^{2}}} + log | \sqrt{1 - x^{2}} | + C .$

22. $\int \frac{(\cos 5 x + \cos 4 x)}{1 - 2 \cos 3 x} d x$

Ans: Let $I = \int \frac{(\cos 5 x + \cos 4 x)}{1 - 2 \cos 3 x} d x$ .

Applying $[\cos C + \cos D = 2 \cos \frac{C + D}{2} . \cos \frac{C - D}{2}$ d $\cos 2 x = 2 c o s^{2} x - 1]$

⇒ $I = \int \frac{2 \cos \frac{9 x}{2} . \cos \frac{x}{2}}{1 - 2 (2 co s^{2} \frac{3 x}{2} - 1)} d x$

⇒ $I = \int \frac{2 \cos \frac{9 x}{2} . \cos \frac{x}{2}}{1 - 4 co s^{2} \frac{3 x}{2} + 2} d x$

⇒ $I = \int \frac{2 \cos \frac{9 x}{2} . \cos \frac{x}{2}}{3 - 4 co s^{2} \frac{3 x}{2}} d x$

Now, multiply and divide by $\cos \frac{3 x}{2}$ , we get

⇒ $I = \int \frac{2 \cos \frac{9 x}{2} . \cos \frac{x}{2} . \cos \frac{3 x}{2}}{3 \cos \frac{3 x}{2} - 4 co s^{3} \frac{3 x}{2}} d x$

⇒ $I = - \int \frac{2 \cos \frac{9 x}{2} . \cos \frac{x}{2} . \cos \frac{3 x}{2}}{4 co s^{3} \frac{3 x}{2} - 3 \cos \frac{3 x}{2}} d x$

⇒ $I = - \int \frac{2 \cos \frac{9 x}{2} . \cos \frac{x}{2} . \cos \frac{3 x}{2}}{\cos 3. \frac{3 x}{2}} d x$

⇒ $I = - \int \frac{2 \cos \frac{9 x}{2} . \cos \frac{x}{2} . \cos \frac{3 x}{2}}{\cos \frac{9 x}{2}} d x$

⇒ $I = - \int 2 \cos \frac{x}{2} . \cos \frac{3 x}{2} d x$

⇒ $I = - \int [\cos (\frac{3 x}{2} + \frac{x}{2}) + \cos (\frac{3 x}{2} - \frac{x}{2})] d x$

⇒ $I = - \int (\cos 2 x + \cos x) d x$

⇒ $I = - \int \cos 2 x d x - \int \cos x d x$

⇒ $I = - (\frac{\sin 2 x}{2}) - \sin x + C$

Hence, $\int \frac{(\cos 5 x + \cos 4 x)}{1 - 2 \cos 3 x} d x = - (\frac{\sin 2 x}{2}) - \sin x + C .$

23. $\int \frac{s i n^{6} x + c o s^{6} x}{s i n^{2} x . c o s^{2} x} d x$

Ans: Let $I = \int \frac{s i n^{6} x + c o s^{6} x}{s i n^{2} x . c o s^{2} x} d x$

⇒ $I = \int \frac{{(s i n^{2} x)}^{3} + {(c o s^{2} x)}^{3}}{s i n^{2} x . c o s^{2} x} d x$

⇒ $I = \int \frac{(s i n^{2} x + c o s^{2} x) (s i n^{4} x - s i n^{2} x . c o s^{2} x + c o s^{4} x)}{s i n^{2} x . c o s^{2} x} d x$

⇒ $I = \int \frac{(s i n^{4} x - s i n^{2} x . c o s^{2} x + c o s^{4} x)}{s i n^{2} x . c o s^{2} x} d x$

⇒ $I = \int \frac{s i n^{4} x}{s i n^{2} x . c o s^{2} x} d x - \int \frac{s i n^{2} x . c o s^{2} x}{s i n^{2} x . c o s^{2} x} d x + \int \frac{c o s^{4} x}{s i n^{2} x . c o s^{2} x} d x$

⇒ $I = \int \frac{s i n^{2} x}{c o s^{2} x} d x - \int d x + \int \frac{c o s^{2} x}{s i n^{2} x} d x$

⇒ $I = \int t a n^{2} x d x - \int d x + \int c o t^{2} x d x$

⇒ $I = \int (s e c^{2} x - 1) d x - \int d x + \int (c o s e c^{2} x - 1) d x$

⇒ $I = \int (s e c^{2} x d x + \int c o s e c^{2} x d x - 3 \int d x$

⇒ $I = t a n x - c o t x - 3 x + C$

Hence, $\int \frac{s i n^{6} x + c o s^{6} x}{s i n^{2} x . c o s^{2} x} d x = t a n x - c o t x - 3 x + C .$

24. $\int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} d x$

Ans: Let $I = \int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} d x$

⇒ $I = \int \frac{x^{\frac{1}{2}}}{\sqrt{{(a^{\frac{3}{2}})}^{2} - {(x^{\frac{3}{2}})}^{2}}} d x$

Here, Put $x^{\frac{3}{2}} = t$

⇒ $\frac{3}{2} x^{\frac{1}{2}} d x = d t$

⇒ $x^{\frac{1}{2}} d x = \frac{2}{3} d t$

Therefore,

⇒ $I = \frac{2}{3} \int \frac{d t}{\sqrt{{(a^{\frac{3}{2}})}^{2} - {(t)}^{2}}}$

⇒ $I = \frac{2}{3} s i n^{- 1} (\frac{t}{a^{\frac{3}{2}}}) + C$

⇒ $I = \frac{2}{3} s i n^{- 1} (\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}) + C$

⇒ $I = \frac{2}{3} s i n^{- 1} {(\frac{x}{a})}^{\frac{3}{2}} + C$

Hence, $\int \frac{\sqrt{x}}{\sqrt{a^{3} - x^{3}}} d x = \frac{2}{3} s i n^{- 1} {(\frac{x}{a})}^{\frac{3}{2}} + C .$

25. $\int \frac{c o s x - c o s 2 x}{1 - c o s x} d x$

Ans: Let $I = \int \frac{c o s x - c o s 2 x}{1 - c o s x} d x$

⇒ $I = \int \frac{2 s i n \frac{3 x}{2} . s i n \frac{x}{2}}{2 s i n^{2} \frac{x}{2}} d x$

⇒ $I = \int \frac{s i n \frac{3 x}{2}}{s i n \frac{x}{2}} d x$

⇒ $I = \int \frac{3 s i n \frac{x}{2} - 4 s i n^{3} \frac{x}{2}}{s i n \frac{x}{2}} d x$

⇒ $I = 3 \int d x - 4 \int s i n^{2} \frac{x}{2} d x$

⇒ $I = 3 x - 4 \int \frac{1 + c o s x}{2} d x$

⇒ $I = 3 x - 2 \int (1 - c o s x) d x$

⇒ $I = 3 x - 2 (x - s i n x) + C$

⇒ $I = 3 x - 2 x + 2 s i n x + C$

⇒ $I = x + 2 s i n x + C$

Hence, $\int \frac{c o s x - c o s 2 x}{1 - c o s x} d x = x + 2 s i n x + C .$

26. $\int \frac{d x}{x \sqrt{x^{4} - 1}}$

Ans: Let $I = \int \frac{d x}{x \sqrt{x^{4} - 1}}$

Multiply and divide by $x^{3}$ , we get

⇒ $I = \int \frac{x^{3} d x}{x^{4} \sqrt{x^{4} - 1}}$

Now, put $x^{4} - 1 = t^{2}$

⇒ $4 x^{3} d x = 2 t d t$

⇒ $x^{3} d x = \frac{t}{2} d t$

Therefore,

⇒ $I = \frac{1}{2} \int \frac{t d t}{(1 + t^{2}) t}$

⇒ $I = \frac{1}{2} \int \frac{d t}{(1 + t^{2})}$

⇒ $I = \frac{1}{2} t a n^{- 1} t + C$

⇒ $I = \frac{1}{2} t a n^{- 1} (\sqrt{x^{4} - 1}) + C$

Hence, $\int \frac{d x}{x \sqrt{x^{4} - 1}} = \frac{1}{2} t a n^{- 1} (\sqrt{x^{4} - 1}) + C .$

Evaluate the Following As Limit of Sums:

27. $\underset{0}{\int^{2}} (x^{2} + 3) d x$

Ans: Let $I = \underset{0}{\int^{2}} (x^{2} + 3) d x$

Here, we have $a = 0, b = 2$ and $h = \frac{b - a}{n} = \frac{2}{n}$

⇒ $n h = 2$ and $f (x) = (x^{2} + 3)$

Now, $\underset{0}{\int^{2}} (x^{2} + 3) d x = lim_{h \to 0} h [f (0) + f (0 + h) + \dots + f {0 + (n - 1) h}]$

$= lim_{h \to 0} h [3 + h^{2} + 3 + 2^{2} h^{2} + 3 + \dots + {(n - 1)}^{2} h^{2} + 3]$

$= lim_{h \to 0} h [3 n + h^{2} {1^{2} + 2^{2} + 3^{2} + \dots + {(n - 1)}^{2}}]$

$= lim_{h \to 0} h [3 n + h^{2} \times \frac{n (n - 1) (2 n - 1)}{6}]$

$= lim_{h \to 0} [3 n h + h^{3} \times \frac{n (n - 1) (2 n - 1)}{6}]$

$= lim_{h \to 0} [3 n h + \frac{n h (n h - h) (2 n h - h)}{6}]$

$= lim_{h \to 0} [3.2 + \frac{2 (2 - h) (2.2 - h)}{6}]$

$= 6 + \frac{4 \times 4}{6}$

$= 6 + \frac{16}{6}$

$= 6 + \frac{8}{3}$

$= \frac{26}{3}$

Hence, $I = \underset{0}{\int^{2}} (x^{2} + 3) d x = \frac{26}{3}$

28. $\underset{0}{\int^{2}} e^{x} d x$

Ans: let $I = \underset{0}{\int^{2}} e^{x} d x$

Here, we have $a = 0, b = 2$ and $h = \frac{b - a}{n} = \frac{2}{n}$

⇒ $n h = 2$ and $f (x) = e^{x}$

Now, $\underset{0}{\int^{2}} e^{x} d x = lim_{h \to 0} h [f (0) + f (0 + h) + \dots + f {0 + (n - 1) h}]$

$= lim_{h \to 0} h [1 + e^{h} + e^{2 h} + e^{3 h} + \dots + e^{(n - 1) h}]$

$= lim_{h \to 0} h [\frac{1. {(e^{h})}^{n} - 1}{e^{h} - 1}]$

$= lim_{h \to 0} h [\frac{e^{n h} - 1}{e^{h} - 1}]$

$= lim_{h \to 0} h [\frac{e^{2} - 1}{e^{h} - 1}]$

$= e^{2} lim_{h \to 0} \frac{h}{e^{h} - 1} - lim_{h \to 0} \frac{h}{e^{h} - 1}$

$= e^{2} - 1$

Hence, $I = \underset{0}{\int^{2}} e^{x} d x = e^{2} - 1$ .

29. $\underset{0}{\int^{\frac{π}{2}}} \frac{\tan x d x}{1 + m^{2} t a n^{2} x}$

Ans: Let $I = \underset{0}{\int^{\frac{π}{2}}} \frac{\tan x d x}{1 + m^{2} t a n^{2} x}$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{\frac{s i n x}{c o s x} d x}{1 + m^{2} \frac{s i n^{2} x}{c o s^{2} x}}$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{\frac{s i n x}{c o s x} d x}{\frac{c o s^{2} x + m^{2} s i n^{2} x}{c o s^{2} x}}$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{s i n x . c o s x d x}{c o s^{2} x + m^{2} s i n^{2} x}$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{s i n x . c o s x d x}{1 - s i n^{2} x (1 - m^{2})}$

Now, put $s i n^{2} x = t$

⇒ $2. \sin x . \cos x d x = d t$

Here, at $x = 0 \Rightarrow t = 0$ and at $x = \frac{π}{2} \Rightarrow t = 1$

Therefore,

⇒ $I = \frac{1}{2} \underset{0}{\int^{1}} \frac{d t}{1 - t (1 - m^{2})}$

⇒ $I = \frac{1}{2} {[- \log | 1 - t (1 - m^{2}) | \times \frac{1}{1 - m^{2}}]}_{0}^{1}$

⇒ $I = - \frac{1}{2} \times \frac{1}{1 - m^{2}} {[\log | 1 - t (1 - m^{2}) |]}_{0}^{1}$

⇒ $I = - \frac{1}{2 (1 - m^{2})} [\log | 1 - 1 (1 - m^{2}) | - \log | 1 - 0 (1 - m^{2}) |]$

⇒ $I = - \frac{1}{2 (1 - m^{2})} [\log | 1 - 1 + m^{2} | - \log | 1 |]$

⇒ $I = - \frac{1}{2 (1 - m^{2})} [\log | m^{2} |]$

⇒ $I = - \frac{1}{2 (1 - m^{2})} 2 \times \log | m |$

⇒ $I = - \frac{1}{(1 - m^{2})} \log | m |$

Hence, $\underset{0}{\int^{\frac{π}{2}}} \frac{\tan x d x}{1 + m^{2} t a n^{2} x} = - \frac{1}{(1 - m^{2})} \log | m |$ .

30. $\underset{1}{\int^{2}} \frac{d x}{\sqrt{(x - 1) (2 - x)}}$

Ans: Let $I = \underset{1}{\int^{2}} \frac{d x}{\sqrt{(x - 1) (2 - x)}}$

⇒ $I = \underset{1}{\int^{2}} \frac{d x}{\sqrt{2 x - x^{2} - 2 + x}}$

⇒ $I = \underset{1}{\int^{2}} \frac{d x}{\sqrt{- x^{2} + 3 x - 2}}$

⇒ $I = \underset{1}{\int^{2}} \frac{d x}{\sqrt{- (x^{2} - 3 x + 2)}}$

⇒ $I = \underset{1}{\int^{2}} \frac{d x}{\sqrt{- [x^{2} - 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + 2]}}$

⇒ $I = \underset{1}{\int^{2}} \frac{d x}{\sqrt{- [{x^{2} - 3 x + {(\frac{3}{2})}^{2}} - {\frac{9}{4} - 2}]}}$

⇒ $I = \underset{1}{\int^{2}} \frac{d x}{\sqrt{- [{(x - \frac{3}{2})}^{2} - {(\frac{1}{2})}^{2}]}}$

⇒ $I = \underset{1}{\int^{2}} \frac{d x}{\sqrt{[{(\frac{1}{2})}^{2} - {(x - \frac{3}{2})}^{2}]}}$

⇒ $I = {[s i n^{- 1} (\frac{x - \frac{3}{2}}{\frac{1}{2}})]}_{1}^{2}$

⇒ $I = {[s i n^{- 1} (2 x - 3)]}_{1}^{2}$

⇒ $I = [s i n^{- 1} (4 - 3) - s i n^{- 1} (2 - 3)]$

⇒ $I = [s i n^{- 1} (1) - s i n^{- 1} (- 1)]$

⇒ $I = [\frac{π}{2} - (- \frac{π}{2})]$

⇒ $I = [\frac{π}{2} + \frac{π}{2}]$

⇒ $I = π$

Hence, $\underset{1}{\int^{2}} \frac{d x}{\sqrt{(x - 1) (2 - x)}} = π$

Evaluate the Following:

31. $\underset{0}{\int^{1}} \frac{d x}{e^{x} + e^{- x}}$

Ans: Let $I = \underset{0}{\int^{1}} \frac{d x}{e^{x} + e^{- x}}$

⇒ $I = \underset{0}{\int^{1}} \frac{d x}{e^{x} + \frac{1}{e^{x}}}$

⇒ $I = \underset{0}{\int^{1}} \frac{e^{x} . d x}{{(e^{x})}^{2} + 1}$

Now, put $e^{x} = t$

⇒ $e^{x} d x = d t$

Now, at $x = 0 \Rightarrow t = 1$ and at $x = 1 \Rightarrow t = e$

Therefore,

⇒ $I = \underset{1}{\int^{e}} \frac{d t}{t^{2} + 1}$

⇒ $I = {[t a n^{- 1} t]}_{1}^{e}$

⇒ $I = (t a n^{- 1} e - t a n^{- 1} 1)$

⇒ $I = t a n^{- 1} e - \frac{π}{4}$

Hence, $\underset{0}{\int^{1}} \frac{d x}{e^{x} + e^{- x}} = t a n^{- 1} e - \frac{π}{4}$

32. $\underset{0}{\int^{1}} \frac{x d x}{\sqrt{1 + x^{2}}}$

Ans: Let $I = \underset{0}{\int^{1}} \frac{x d x}{\sqrt{1 + x^{2}}}$

Now, put $1 + x^{2} = t^{2}$

⇒ $2 x d x = 2 t d t$

⇒ $x d x = t d t$

here, at $x = 0 \Rightarrow t = 1$ and at $x = 1 \Rightarrow t = \sqrt{2}$

Therefore,

⇒ $I = \underset{1}{\int^{\sqrt{2}}} \frac{t d t}{\sqrt{t^{2}}}$

⇒ $I = \underset{1}{\int^{\sqrt{2}}} \frac{t d t}{t}$

⇒ $I = \underset{1}{\int^{\sqrt{2}}} d t$

⇒ $I = {[t]}_{1}^{\sqrt{2}}$

⇒ $I = (\sqrt{2} - 1)$

Hence, $\underset{0}{\int^{1}} \frac{x d x}{\sqrt{1 + x^{2}}} = (\sqrt{2} - 1)$

33. $\underset{0}{\int^{π}} x \sin x c o s^{2} x d x$

Ans: Let $I = \underset{0}{\int^{π}} x \sin x c o s^{2} x d x$ ………. (i)

⇒ $I = \underset{0}{\int^{π}} (π - x) \sin (π - x) c o s^{2} (π - x) d x$

⇒ $I = \underset{0}{\int^{π}} (π - x) \sin x c o s^{2} x d x$ …………(ii)

Adding eq(i) and (ii), we get

⇒ $2 I = π \underset{0}{\int^{π}} \sin x c o s^{2} x d x$

Now, put $\cos x = t$

⇒ $- \sin x d x = d t$

⇒ $\sin x d x = - d t$

here, at $x = 0 \Rightarrow t = 1$ and at $x = π \Rightarrow t = - 1$

Therefore,

⇒ $2 I = - π \underset{1}{\int^{- 1}} t^{2} d t$

⇒ $2 I = - π {[\frac{t^{3}}{3}]}_{1}^{- 1}$

⇒ $2 I = - \frac{1}{3} π {[t^{3}]}_{1}^{- 1}$

⇒ $2 I = - \frac{1}{3} π (- 1 - 1)$

⇒ 2 $I = \frac{2 π}{3}$

⇒ $I = \frac{π}{3}$

Hence, $\underset{0}{\int^{π}} x \sin x c o s^{2} x d x = \frac{π}{3}$ .

34. $\underset{0}{\int^{\frac{1}{2}}} \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}}$

Ans: Let $I = \underset{0}{\int^{\frac{1}{2}}} \frac{d x}{(1 + x^{2}) \sqrt{1 - x^{2}}}$

Put $x = \sin θ$

⇒ $d x = \cos θ d θ$

Here, at $x = 0 \Rightarrow θ = 0$ and at $x = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$

Therefore,

⇒ $I = \underset{0}{\int^{\frac{π}{6}}} \frac{c o s θ d θ}{(1 + s i n^{2} θ) \sqrt{1 - s i n^{2} θ}}$

⇒ $I = \underset{0}{\int^{\frac{π}{6}}} \frac{c o s θ d θ}{(1 + s i n^{2} θ) c o s θ}$

⇒ $I = \underset{0}{\int^{\frac{π}{6}}} \frac{d θ}{(1 + s i n^{2} θ)}$

Divide by $c o s^{2} θ$ in numerator and denominator

⇒ $I = \underset{0}{\int^{\frac{π}{6}}} \frac{\frac{1}{c o s^{2} θ} d θ}{(\frac{1}{c o s^{2} θ} + \frac{s i n^{2} θ}{c o s^{2} θ})}$

⇒ $I = \underset{0}{\int^{\frac{π}{6}}} \frac{s e c^{2} θ d θ}{(s e c^{2} θ + t a n^{2} θ)}$

⇒ $I = \underset{0}{\int^{\frac{π}{6}}} \frac{s e c^{2} θ d θ}{(1 + t a n^{2} θ + t a n^{2} θ)}$

⇒ $I = \underset{0}{\int^{\frac{π}{6}}} \frac{s e c^{2} θ d θ}{(1 + 2 t a n^{2} θ)}$

Now, put $\tan θ = t$

⇒ $s e c^{2} θ d θ = d t$

Here, at $θ = 0 \Rightarrow t = 0$ and at $θ = \frac{π}{6} \Rightarrow t = \frac{1}{\sqrt{3}}$

Therefore,

⇒ $I = \underset{0}{\int^{\frac{1}{\sqrt{3}}}} \frac{d t}{(1 + 2 t^{2})}$

⇒ $I = \frac{1}{2} \underset{0}{\int^{\frac{1}{\sqrt{3}}}} \frac{d t}{(\frac{1}{2} + t^{2})}$

⇒ $I = \frac{1}{2} \underset{0}{\int^{\frac{1}{\sqrt{3}}}} \frac{d t}{[{(\frac{1}{\sqrt{2}})}^{2} + t^{2}]}$

⇒ $I = \frac{1}{2} {[\frac{1}{\frac{1}{\sqrt{2}}} t a n^{- 1} (\frac{t}{\frac{1}{\sqrt{2}}})]}_{0}^{\frac{1}{\sqrt{3}}}$

⇒ $I = \frac{1}{\sqrt{2}} {[t a n^{- 1} (\sqrt{2} \times t)]}_{0}^{\frac{1}{\sqrt{3}}}$

⇒ $I = \frac{1}{\sqrt{2}} [t a n^{- 1} (\sqrt{2} \times \frac{1}{\sqrt{3}}) - t a n^{- 1} (0)]$

⇒ $I = \frac{1}{\sqrt{2}} t a n^{- 1} (\frac{\sqrt{2}}{\sqrt{3}})$

Long Answer (L.A)

35. $\int \frac{x^{2} d x}{x^{4} - x^{2} - 12}$

Ans: Let $I = \int \frac{x^{2} d x}{x^{4} - x^{2} - 12}$

⇒ $I = \int \frac{x^{2} d x}{x^{4} - 4 x^{2} + 3 x^{2} - 12}$

⇒ $I = \int \frac{x^{2} d x}{x^{2} (x^{2} - 4) + 3 (x^{2} - 4)}$

⇒ $I = \int \frac{x^{2} d x}{(x^{2} - 4) (x^{2} + 3)}$

⇒ $I = \int \frac{x^{2} - 4 + 4 d x}{(x^{2} - 4) (x^{2} + 3)}$

$\Rightarrow I = \int \frac{(x^{2} - 4) d x}{(x^{2} - 4) (x^{2} + 3)} + 4 \int \frac{d x}{(x^{2} - 4) (x^{2} + 3)}$

$\Rightarrow I = \int \frac{d x}{(x^{2} + 3)} + 4 \int \frac{[(x^{2} - 4) - (x^{2} + 3)} d x}{(- 7) (x^{2} - 4) (x^{2} + 3)}$

$\Rightarrow I = \int \frac{d x}{(x^{2} + 3)} - \frac{4}{7} \int \frac{(x^{2} - 4) d x}{(x^{2} - 4) (x^{2} + 3)} + \frac{4}{7} \int \frac{(x^{2} + 3) d x}{(x^{2} - 4) (x^{2} + 3)}$

$\Rightarrow I = \int \frac{d x}{(x^{2} + 3)} - \frac{4}{7} \int \frac{d x}{(x^{2} + 3)} + \frac{4}{7} \int \frac{d x}{(x^{2} - 4)}$

$\Rightarrow I = \frac{3}{7} \int \frac{d x}{(x^{2} + 3)} + \frac{4}{7} \int \frac{d x}{(x^{2} - 4)}$

$\Rightarrow I = \frac{3}{7} \int \frac{d x}{(x^{2} + {(\sqrt{3})}^{2})} + \frac{4}{7} \int \frac{d x}{(x^{2} - 2^{2})}$

$\Rightarrow I = \frac{3}{7} [\frac{1}{\sqrt{3}} t a n^{- 1} \frac{x}{\sqrt{3}}] + \frac{4}{7} . \frac{1}{2 \times 2} \log | \frac{x - 2}{x + 2} | + C$

$\Rightarrow I = \frac{\sqrt{3}}{7} t a n^{- 1} \frac{x}{\sqrt{3}} + \frac{1}{7} \log | \frac{x - 2}{x + 2} | + C$

Hence, $\int \frac{x^{2} d x}{x^{4} - x^{2} - 12} = \frac{\sqrt{3}}{7} t a n^{- 1} \frac{x}{\sqrt{3}} + \frac{1}{7} \log | \frac{x - 2}{x + 2} | + C .$

36. $\int \frac{x^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

Ans: Let $I = \int \frac{x^{2} + a^{2} - a^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

⇒ $I = \int \frac{(x^{2} + a^{2}) d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} - \int \frac{a^{2} d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

⇒ $I = \int \frac{d x}{(x^{2} + b^{2})} - a^{2} \int \frac{[(x^{2} + a^{2}) - (x^{2} + b^{2})] d x}{(a^{2} - b^{2}) (x^{2} + a^{2}) (x^{2} + b^{2})}$

⇒ $I = \int \frac{d x}{(x^{2} + b^{2})} - \frac{a^{2}}{(a^{2} - b^{2})} \int \frac{[(x^{2} + a^{2}) - (x^{2} + b^{2})] d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

⇒ $I = \int \frac{d x}{(x^{2} + b^{2})} - \frac{a^{2}}{(a^{2} - b^{2})} [\int \frac{(x^{2} + a^{2}) d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} - \int \frac{(x^{2} + b^{2}) d x}{(x^{2} + a^{2}) (x^{2} + b^{2})}]$

⇒ $I = \int \frac{d x}{(x^{2} + b^{2})} - \frac{a^{2}}{(a^{2} - b^{2})} [\int \frac{d x}{(x^{2} + b^{2})} - \int \frac{d x}{(x^{2} + a^{2})}]$

⇒ $I = \frac{- b^{2}}{(a^{2} - b^{2})} \int \frac{d x}{(x^{2} + b^{2})} + \frac{a^{2}}{(a^{2} - b^{2})} \int \frac{d x}{(x^{2} + a^{2})}$

⇒ $I = \frac{b^{2}}{(b^{2} - a^{2})} \frac{1}{b} t a n^{- 1} \frac{x}{b} - \frac{a^{2}}{(b^{2} - a^{2})} \frac{1}{a} t a n^{- 1} \frac{x}{a} + C$

⇒ $I = \frac{b}{(b^{2} - a^{2})} t a n^{- 1} \frac{x}{b} - \frac{a}{(b^{2} - a^{2})} t a n^{- 1} \frac{x}{a} + C$

37. $\underset{0}{\int^{π}} \frac{x}{1 + \sin x} d x$

Ans: Let $I = \underset{0}{\int^{π}} \frac{x}{1 + \sin x} d x$ ……….(i)

⇒ $I = \underset{0}{\int^{π}} \frac{(π - x)}{1 + \sin (π - x)} d x$

⇒ $I = \underset{0}{\int^{π}} \frac{(π - x)}{1 + \sin x} d x$ ………….(ii)

Now, adding eq(i) and (ii), we get

$\Rightarrow 2 I = \underset{0}{\int^{π}} \frac{π}{1 + \sin x} d x$

$\Rightarrow 2 I = π \underset{0}{\int^{π}} \frac{(1 - \sin x)}{(1 + \sin x) (1 - \sin x)} d x$

$\Rightarrow 2 I = π \underset{0}{\int^{π}} \frac{(1 - \sin x)}{(1 - s i n^{2} x)} d x$

$\Rightarrow 2 I = π \underset{0}{\int^{π}} \frac{(1 - \sin x)}{c o s^{2} x} d x$

$\Rightarrow 2 I = π \underset{0}{\int^{π}} (\frac{1}{c o s^{2} x} - \frac{\sin x}{c o s^{2} x}) d x$

$\Rightarrow 2 I = π \underset{0}{\int^{π}} (s e c^{2} x - \sec x . \tan x) d x$

$\Rightarrow 2 I = π {[\tan x - \sec x]}_{0}^{π}$

$\Rightarrow 2 I = π {[\tan π - \sec π - (\tan 0 - \sec 0)]}_{0}^{π}$

$\Rightarrow 2 I = π [0 - (- 1) - (0 - 1)]$

$\Rightarrow 2 I = π [1 + 1]$

$\Rightarrow 2 I = 2 π$

$\Rightarrow I = π$

Hence, $\underset{0}{\int^{π}} \frac{x}{1 + \sin x} d x = π$

38. $\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x$

Ans: Let $I = \int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x$

Now, $\frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} = \frac{A}{(x - 1)} + \frac{B}{(x + 2)} + \frac{C}{(x - 3)}$

⇒ $2 x - 1 = A (x + 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x + 2)$

Here, put $x = 1$ in above equation, we get

⇒ $1 = A (1 + 2) (1 - 3) + 0 + 0$

⇒ $1 = - 6 A$

⇒ $A = - \frac{1}{6}$

Now, put $x = - 2$

⇒ $2 (- 2) - 1 = 0 + B (- 2 - 1) (- 2 - 3) + 0$

⇒ $- 5 = B (- 3) (- 5)$

⇒ $B = - \frac{1}{3}$

Now, put $x = 3$ , we get

⇒ $2 (3) - 1 = 0 + 0 + C (3 - 1) (3 + 2)$

⇒ $5 = C (2) (5)$

⇒ $C = \frac{1}{2}$

Therefore,

$\Rightarrow \int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x = - \int \frac{d x}{6 (x - 1)} - \int \frac{d x}{3 (x + 2)} + \int \frac{d x}{2 (x - 3)}$

$= - \frac{1}{6} \log | (x - 1) | - \frac{1}{3} \log | (x + 2) | + \frac{1}{2} \log | (x - 3) | + C$

$= - \log {(x - 1)}^{\frac{1}{6}} - \log {(x + 2)}^{\frac{1}{3}} + \log {(x - 3)}^{\frac{1}{2}} + C$

$= \log | \frac{\sqrt{(x - 3)}}{{(x - 1)}^{\frac{1}{6}} {(x + 2)}^{\frac{1}{3}}} | + C$

Hence, $\int \frac{2 x - 1}{(x - 1) (x + 2) (x - 3)} d x = \log | \frac{\sqrt{(x - 3)}}{{(x - 1)}^{\frac{1}{6}} {(x + 2)}^{\frac{1}{3}}} | + C$

39. $\int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$

Ans: Let $I = \int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$

⇒ $I = \int e^{t a n^{- 1} x} (\frac{1 + x^{2}}{1 + x^{2}} + \frac{x}{1 + x^{2}}) d x$

⇒ $I = \int e^{t a n^{- 1} x} (1 + \frac{x}{1 + x^{2}}) d x$

⇒ $I = \int e^{t a n^{- 1} x} d x + \int \frac{x e^{t a n^{- 1} x}}{1 + x^{2}} d x$

⇒ $I = I_{1} + I_{2}$ …...(i), where, $I_{1} = \int e^{t a n^{- 1} x} d x & I_{2} = \int \frac{x e^{t a n^{- 1} x}}{1 + x^{2}} d x$

Now, $I_{2} = \int \frac{x e^{t a n^{- 1} x}}{1 + x^{2}} d x$

Put $t a n^{- 1} x = t$

⇒ $\frac{1}{1 + x^{2}} d x = d t$

Therefore,

⇒ $I_{2} = \int e^{t} . \tan t d t$

⇒ $I_{2} = \tan t \int e^{t} d t - \int (\frac{d}{d x} \tan t . \int e^{t} d t) d t$

⇒ $I_{2} = \tan t . e^{t} - \int s e c^{2} t . e^{t} d t + C$

⇒ $I_{2} = x . e^{t a n^{- 1} x} - \int (1 + t a n^{2} t) . e^{t a n^{- 1} x} \frac{1}{1 + x^{2}} d x + C$

⇒ $I_{2} = x . e^{t a n^{- 1} x} - \int (1 + x^{2}) . e^{t a n^{- 1} x} \frac{1}{1 + x^{2}} d x + C$

⇒ $I_{2} = x . e^{t a n^{- 1} x} - \int e^{t a n^{- 1} x} d x + C$

Put values of $I_{1}$ and $I_{2}$ , in equation (i),

⇒ $I = \int e^{t a n^{- 1} x} d x + x . e^{t a n^{- 1} x} - \int e^{t a n^{- 1} x} d x + C$

⇒ $I = x . e^{t a n^{- 1} x} + C$

Hence, value of $\int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x = x . e^{t a n^{- 1} x} + C .$

40. $\int s i n^{- 1} \sqrt{\frac{x}{a + x}} d x$

Ans: Let $I = \int s i n^{- 1} \sqrt{\frac{x}{a + x}} d x$

Put $x = a t a n^{2} θ$

⇒ $d x = a . 2 \tan θ . s e c^{2} θ d θ$

Therefore,

$\Rightarrow I = \int s i n^{- 1} \sqrt{\frac{a t a n^{2} θ}{a + a t a n^{2} θ}} \times$ $2 a \tan θ . s e c^{2} θ d θ$

$\Rightarrow I = \int s i n^{- 1} \sqrt{\frac{a t a n^{2} θ}{a (1 + t a n^{2} θ)}} \times$ $2 a \tan θ . s e c^{2} θ d θ$

$\Rightarrow I = \int s i n^{- 1} \sqrt{\frac{t a n^{2} θ}{(1 + t a n^{2} θ)}} \times$ $2 a \tan θ . s e c^{2} θ d θ$

$\Rightarrow I = \int s i n^{- 1} \sqrt{\frac{t a n^{2} θ}{s e c^{2} θ}} \times$ $2 a \tan θ . s e c^{2} θ d θ$

$\Rightarrow I = \int s i n^{- 1} \sqrt{s i n^{2} θ} \times$ $2 a \tan θ . s e c^{2} θ d θ$

$\Rightarrow I = \int s i n^{- 1} (\sin θ) \times$ $2 a \tan θ . s e c^{2} θ d θ$

$\Rightarrow I = 2 a \int θ \tan θ . s e c^{2} θ d θ$

$\Rightarrow I = 2 a [θ \int \tan θ . s e c^{2} θ d θ - \int (\frac{d}{d θ} θ . \int \tan θ . s e c^{2} θ d θ) d θ]$

Put $\tan θ = t$

$\Rightarrow$ $s e c^{2} θ d θ = d t$

$\Rightarrow I = 2 a [θ \int t d t - \int (\int t d t) d θ]$

$\Rightarrow I = 2 a [θ . \frac{t^{2}}{2} - \int \frac{t^{2}}{2} d θ]$

$\Rightarrow I = 2 a [θ . \frac{t a n^{2} θ}{2} - \int \frac{t a n^{2} θ}{2} d θ]$

$\Rightarrow I = a [θ (t a n^{2} θ) - \int t a n^{2} θ d θ]$

$\Rightarrow I = a θ (t a n^{2} θ) -$ $a \int (s e c^{2} θ - 1) d θ$

$\Rightarrow I = a θ (t a n^{2} θ) - a \tan θ + a θ + C$

$\Rightarrow I = a [\frac{x}{a} (t a n^{- 1} \sqrt{\frac{x}{a}}) - \sqrt{\frac{x}{a}} + t a n^{- 1} \sqrt{\frac{x}{a}}] + C$

Hence, $\int s i n^{- 1} \sqrt{\frac{x}{a + x}} d x = a [\frac{x}{a} (t a n^{- 1} \sqrt{\frac{x}{a}}) - \sqrt{\frac{x}{a}} + t a n^{- 1} \sqrt{\frac{x}{a}}] + C$ .

41. $\underset{\frac{π}{3}}{\int^{\frac{π}{2}}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{5}{2}}} d x \dots$

Ans: Let $I = \underset{\frac{π}{3}}{\int^{\frac{π}{2}}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{5}{2}}} d x$

⇒ $I = \underset{\frac{π}{3}}{\int^{\frac{π}{2}}} \frac{\sqrt{2 co s^{2} \frac{x}{2}}}{{(2 s i n^{2} \frac{x}{2})}^{\frac{5}{2}}} d x$

⇒ $I = \frac{\sqrt{2}}{2^{\frac{5}{2}}} \underset{\frac{π}{3}}{\int^{\frac{π}{2}}} \frac{cos \frac{x}{2}}{s i n^{5} \frac{x}{2}} d x$

Put $\sin \frac{x}{2} = t$

⇒ $\frac{1}{2} \cos \frac{x}{2} d x = d t$

⇒ $\cos \frac{x}{2} d x = 2 d t$

Here, at $x = \frac{π}{3} \Rightarrow t = \frac{1}{2}$ and at $x = \frac{π}{2} \Rightarrow t = \frac{1}{\sqrt{2}}$

Therefore,

⇒ $I = \frac{\sqrt{2}}{\sqrt{32}} \underset{\frac{1}{2}}{\int^{\frac{1}{\sqrt{2}}}} \frac{2 d t}{t^{5}}$

⇒ $I = \frac{\sqrt{2}}{4 \sqrt{2}} \times 2 \underset{\frac{1}{2}}{\int^{\frac{1}{\sqrt{2}}}} t^{- 5} d t$

⇒ $I = \frac{1}{2} {[\frac{t^{- 4}}{- 4}]}_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}$

⇒ $I = - \frac{1}{8} {[t^{- 4}]}_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}$

⇒ $I = - \frac{1}{8} [{(\frac{1}{\sqrt{2}})}^{- 4} - {(\frac{1}{2})}^{- 4}]$

⇒ $I = - \frac{1}{8} [{(\sqrt{2})}^{4} - {(2)}^{4}]$

⇒ $I = - \frac{1}{8} [4 - 16]$

⇒ $I = \frac{1}{8} \times 12$

⇒ $I = \frac{3}{2}$

Hence, $\underset{\frac{π}{3}}{\int^{\frac{π}{2}}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{5}{2}}} d x = \frac{3}{2} .$

42. $\int e^{- 3 x} c o s^{3} x d x$

Ans: Let $I = \int e^{- 3 x} c o s^{3} x d x$

⇒ $I = \int e^{- 3 x} (\frac{\cos 3 x + 3 \cos x}{4}) d x$

⇒ $I = \frac{1}{4} \int (e^{- 3 x} \cos 3 x + 3 e^{- 3 x} \cos x) d x$

⇒ $I = \frac{1}{4} \int e^{- 3 x} \cos 3 x d x + \frac{3}{4} \int e^{- 3 x} \cos x d x$

⇒ $I = \frac{1}{4} I_{1} + \frac{3}{4} I_{2}$ ………… eq (i)

Where, $I_{1} = \int e^{- 3 x} \cos 3 x d x$ and $I_{2} = \int 3 e^{- 3 x} \cos x d x$

Now, $I_{1} = \int e^{- 3 x} \cos 3 x d x$

⇒ $I_{1} = \cos 3 x \int e^{- 3 x} d x - \int (\frac{d}{d x} \cos 3 x . \int e^{- 3 x} d x) d x$

⇒ $I_{1} = \frac{e^{- 3 x}}{- 3} \cos 3 x - \int (- 3 sin 3 x . \frac{e^{- 3 x}}{- 3}) d x$

⇒ $I_{1} = \frac{e^{- 3 x}}{- 3} \cos 3 x - \int (e^{- 3 x} sin 3 x .) d x$

⇒ $I_{1} = \frac{e^{- 3 x}}{- 3} \cos 3 x - [\sin 3 x \int e^{- 3 x} d x - \int (\frac{d}{d x} \sin 3 x . \int e^{- 3 x} d x) d x]$

⇒ $I_{1} = \frac{e^{- 3 x}}{- 3} \cos 3 x - [\frac{e^{- 3 x}}{- 3} \sin 3 x - \int (3 \cos 3 x . \frac{e^{- 3 x}}{- 3}) d x]$

⇒ $I_{1} = \frac{e^{- 3 x}}{- 3} \cos 3 x - [\frac{e^{- 3 x}}{- 3} \sin 3 x + \int (e^{- 3 x} \cos 3 x) d x]$

⇒ $I_{1} = \frac{e^{- 3 x}}{- 3} \cos 3 x - [\frac{e^{- 3 x}}{- 3} \sin 3 x + I_{1}]$

⇒ $I_{1} = \frac{e^{- 3 x}}{- 3} \cos 3 x + \frac{e^{- 3 x}}{3} \sin 3 x - I_{1}$

⇒ $2 I_{1} = \frac{e^{- 3 x}}{- 3} \cos 3 x + \frac{e^{- 3 x}}{3} \sin 3 x$

⇒ $2 I_{1} = \frac{1}{3} e^{- 3 x} (- \cos 3 x + \sin 3 x) + C_{1}$

⇒ $I_{1} = \frac{1}{6} e^{- 3 x} (\sin 3 x - \cos 3 x) + C_{1}$

Now, $I_{2} = \int e^{- 3 x} \cos x d x$

⇒ $I_{2} = \int e^{- 3 x} \cos x d x$

⇒ $I_{2} = \cos x \int e^{- 3 x} d x - \int (\frac{d}{d x} \cos x . \int e^{- 3 x} d x) d x$

⇒ $I_{2} = \frac{e^{- 3 x}}{- 3} \cos x - \int (- \sin x . \frac{e^{- 3 x}}{- 3}) d x$

⇒ $I_{2} = \frac{e^{- 3 x}}{- 3} \cos x - \frac{1}{3} \int (e^{- 3 x} . \sin x) d x$

⇒ $I_{2} = \frac{e^{- 3 x}}{- 3} \cos x - \frac{1}{3} [\sin x \int e^{- 3 x} d x - \int (\frac{d}{d x} \sin x . \int e^{- 3 x} d x) d x]$

⇒ $I_{2} = \frac{e^{- 3 x}}{- 3} \cos x - \frac{1}{3} [\frac{e^{- 3 x}}{- 3} \sin x - \int (\frac{e^{- 3 x}}{- 3} \cos x) d x]$

⇒ $I_{2} = \frac{e^{- 3 x}}{- 3} \cos x - \frac{1}{3} [\frac{e^{- 3 x}}{- 3} \sin x + \frac{1}{3} \int (e^{- 3 x} \cos x) d x]$

⇒ $I_{2} = \frac{e^{- 3 x}}{- 3} \cos x + \frac{1}{9} e^{- 3 x} \sin x - \frac{1}{9} \int (e^{- 3 x} \cos x) d x$

⇒ $I_{2} = \frac{e^{- 3 x}}{- 3} \cos x + \frac{1}{9} e^{- 3 x} \sin x - \frac{1}{9} I_{2}$

⇒ $I_{2} + \frac{I_{2}}{9} = \frac{e^{- 3 x}}{- 3} \cos x + \frac{1}{9} e^{- 3 x} \sin x$

⇒ $\frac{10 I_{2}}{9} = \frac{- 3 e^{- 3 x} . \cos x + e^{- 3 x} \sin x}{9} + C_{2}$

⇒ $10 I_{2} = e^{- 3 x} (\sin x - 3 \cos x) + C_{2}$

⇒ $I_{2} = \frac{1}{10} e^{- 3 x} (\sin x - 3 \cos x) + C_{2}$

Now, put the value of $I_{1}$ and $I_{2}$ in eq(i), we get

⇒ $I = \frac{1}{24} e^{- 3 x} (\sin 3 x - \cos 3 x) + \frac{3}{40} e^{- 3 x} (\sin x - 3 \cos x) + C$

43. $\int \sqrt{\tan x} d x$

Ans: Let $I = \int \sqrt{\tan x} d x$

Put $\tan x = t^{2}$

⇒ $s e c^{2} x d x = 2 t d t$

⇒ $(1 + t a n^{2} x) d x = 2 t d t$

⇒ $d x = \frac{2 t}{(1 + t a n^{2} x)} d t$

⇒ $d x = \frac{2 t}{(1 + t^{4})} d t$

Therefore,

⇒ $I = \int \sqrt{t^{2}} . \frac{2 t}{(1 + t^{4})} d t$

⇒ $I = \int \frac{2 t^{2}}{(1 + t^{4})} d t$

⇒ $I = \int \frac{(1 + t^{2}) - (1 - t^{2})}{(1 + t^{4})} d t$

⇒ $I = \int \frac{(1 + t^{2})}{(1 + t^{4})} d t - \int \frac{(1 - t^{2})}{(1 + t^{4})} d t$

⇒ $I = \int \frac{(\frac{1}{t^{2}} + 1)}{(\frac{1}{t^{2}} + t^{2})} d t - \int \frac{(\frac{1}{t^{2}} - 1)}{(\frac{1}{t^{2}} + t^{2})} d t$

⇒ $I = \int \frac{(\frac{1}{t^{2}} + 1)}{{(t - \frac{1}{t})}^{2} + 2} d t - \int \frac{(\frac{1}{t^{2}} - 1)}{{(t + \frac{1}{t})}^{2} - 2} d t$

Put $u = (t - \frac{1}{t}) \Rightarrow d u = (\frac{1}{t^{2}} + 1) d t$

And $v = (t + \frac{1}{t}) \Rightarrow d u = - (\frac{1}{t^{2}} - 1) d t$

Therefore,

⇒ $I = \int \frac{d u}{{(u)}^{2} + 2} - \int \frac{- d v}{{(v)}^{2} - 2}$

⇒ $I = \int \frac{d u}{{(u)}^{2} + {(\sqrt{2})}^{2}} + \int \frac{d v}{{(v)}^{2} - {(\sqrt{2})}^{2}}$

⇒ $I = \frac{1}{\sqrt{2}} t a n^{- 1} \frac{u}{\sqrt{2}} + \frac{1}{2 \sqrt{2}} \log | \frac{v - \sqrt{2}}{v + \sqrt{2}} | + C$

⇒ $I = \frac{1}{\sqrt{2}} t a n^{- 1} \frac{(t - \frac{1}{t})}{\sqrt{2}} + \frac{1}{2 \sqrt{2}} \log | \frac{(t + \frac{1}{t}) - \sqrt{2}}{(t + \frac{1}{t}) + \sqrt{2}} | + C$

⇒ $I = \frac{1}{\sqrt{2}} t a n^{- 1} \frac{(\sqrt{\tan x} - \frac{1}{\sqrt{\tan x}})}{\sqrt{2}} + \frac{1}{2 \sqrt{2}} \log | \frac{(\sqrt{\tan x} + \frac{1}{\sqrt{\tan x}}) - \sqrt{2}}{(\sqrt{\tan x} + \frac{1}{\sqrt{\tan x}}) + \sqrt{2}} | + C$

⇒ $I = \frac{1}{\sqrt{2}} t a n^{- 1} \frac{(\tan x - 1)}{\sqrt{2 \tan x}} + \frac{1}{2 \sqrt{2}} \log | \frac{\tan x - \sqrt{2 \tan x} + 1}{\tan x + \sqrt{2 \tan x} + 1} | + C$

44. $\underset{0}{\int^{\frac{π}{2}}} \frac{d x}{{(a^{2} c o s^{2} x + b^{2} s i n^{2} x)}^{2}}$

Ans: Let $I = \underset{0}{\int^{\frac{π}{2}}} \frac{d x}{{(a^{2} c o s^{2} x + b^{2} s i n^{2} x)}^{2}}$

Divide numerator and denominator by $c o s^{4} x$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{\frac{d x}{c o s^{4} x}}{\frac{{(a^{2} c o s^{2} x + b^{2} s i n^{2} x)}^{2}}{c o s^{4} x}}$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{s e c^{4} x d x}{{(a^{2} + b^{2} t a n^{2} x)}^{2}}$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{(1 + t a n^{2} x) s e c^{2} x d x}{{(a^{2} + b^{2} t a n^{2} x)}^{2}}$

Put $b \tan x = a \tan t$

⇒ $b s e c^{2} x d x = a s e c^{2} t d t$

⇒ $s e c^{2} x d x = \frac{a}{b} s e c^{2} t d t$

When $x \to 0, t \to 0$ and When $x \to \frac{π}{2}, \tan t \to 0, t \to \frac{π}{2}$

Therefore,

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \frac{(1 + {(\frac{a}{b} \tan t)}^{2})}{a^{4} s e c^{4} t} \frac{a}{b} s e c^{2} t d t$

⇒ $I = \frac{1}{a^{3} b^{3}} \underset{0}{\int^{\frac{π}{2}}} (b^{2} + a^{2} t a n^{2} t) c o s^{2} t d t$

⇒ $I = \frac{1}{a^{3} b^{3}} \underset{0}{\int^{\frac{π}{2}}} (b^{2} c o s^{2} t + a^{2} s i n^{2} t) d t$

⇒ $I = \frac{1}{a^{3} b^{3}} \underset{0}{\int^{\frac{π}{2}}} (b^{2} \frac{1 + \cos 2 t}{2} + a^{2} \frac{1 - \cos 2 t}{2}) d t$

⇒ $I = \frac{1}{2 a^{3} b^{3}} \underset{0}{\int^{\frac{π}{2}}} [(a^{2} + b^{2}) + (b^{2} - a^{2}) \cos 2 t] d t$

⇒ $I = \frac{1}{2 a^{3} b^{3}} {[(a^{2} + b^{2}) x - \frac{1}{2} (b^{2} - a^{2}) \sin 2 t]}_{0}^{\frac{π}{2}}$

⇒ $I = \frac{1}{2 a^{3} b^{3}} [(a^{2} + b^{2}) \frac{π}{2} - \frac{1}{2} (b^{2} - a^{2}) \sin π]$

⇒ $I = \frac{1}{2 a^{3} b^{3}} [(a^{2} + b^{2}) \frac{π}{2}]$

⇒ $I = \frac{π (a^{2} + b^{2})}{4 a^{3} b^{3}}$

Hence, $\underset{0}{\int^{\frac{π}{2}}} \frac{d x}{{(a^{2} c o s^{2} x + b^{2} s i n^{2} x)}^{2}} = \frac{π (a^{2} + b^{2})}{4 a^{3} b^{3}} .$

45. $\underset{0}{\int^{1}} x \log (1 + 2 x) d x$

Ans: Let $I = \underset{0}{\int^{1}} x \log (1 + 2 x) d x$

⇒ $I = {[\log (1 + 2 x) \underset{0}{\int^{1}} x d x]}_{0}^{1} - \underset{0}{\int^{1}} (\frac{d}{d x} \log (1 + 2 x) . \underset{0}{\int^{1}} x d x) d x$

⇒ $I = {[\frac{x^{2}}{2} \log (1 + 2 x)]}_{0}^{1} - \underset{0}{\int^{1}} (\frac{2}{1 + 2 x} . \frac{x^{2}}{2}) d x$

⇒ $I = (\frac{1}{2} \log 3 - 0) - \underset{0}{\int^{1}} (\frac{x^{2}}{1 + 2 x}) d x$

⇒ $I = (\frac{1}{2} \log 3 - 0) - \underset{0}{\int^{1}} (\frac{x}{2} - \frac{\frac{x}{2}}{1 + 2 x}) d x$

⇒ $I = (\frac{1}{2} \log 3) - \frac{1}{2} \underset{0}{\int^{1}} x d x + \frac{1}{2} \underset{0}{\int^{1}} \frac{x}{1 + 2 x} d x$

⇒ $I = \frac{1}{2} \log 3 - \frac{1}{2} (\frac{1}{2} - 0) + \frac{1}{4} \underset{0}{\int^{1}} \frac{1 + 2 x - 1}{1 + 2 x} d x$

⇒ $I = \frac{1}{2} \log 3 - \frac{1}{4} + \frac{1}{4} \underset{0}{\int^{1}} \frac{1 + 2 x}{1 + 2 x} d x - \frac{1}{4} \underset{0}{\int^{1}} \frac{d x}{1 + 2 x}$

⇒ $I = \frac{1}{2} \log 3 - \frac{1}{4} + \frac{1}{4} \underset{0}{\int^{1}} d x - \frac{1}{4} {[\frac{1}{2} \log (1 + 2 x)]}_{0}^{1}$

⇒ $I = \frac{1}{2} \log 3 - \frac{1}{4} + \frac{1}{4} {(x)}_{0}^{1} - \frac{1}{8} [\log (3) - 0]$

⇒ $I = \frac{1}{2} \log 3 - \frac{1}{4} + \frac{1}{4} - \frac{1}{8} [\log (3)]$

⇒ $I = \frac{1}{2} \log 3 - \frac{1}{8} \log 3$

⇒ $I = \frac{3}{8} \log 3$

Hence, $\underset{0}{\int^{1}} x \log (1 + 2 x) d x = \frac{3}{8} \log 3.$

46. $\underset{0}{\int^{π}} x \log \sin x d x$

Ans: Let $I = \underset{0}{\int^{π}} x \log \sin x d x$ ……………….. (i)

⇒ $I = \underset{0}{\int^{π}} (π - x) \log \sin (π - x) d x$

⇒ $I = \underset{0}{\int^{π}} (π - x) \log \sin x d x$ …………..(ii)

Adding eq (i) and (ii), we get

⇒ $2 I = π \underset{0}{\int^{π}} \log \sin x d x$

⇒ $2 I = 2 π \underset{0}{\int^{\frac{π}{2}}} \log \sin x d x$

⇒ $I = π \underset{0}{\int^{\frac{π}{2}}} \log \sin x d x$ ………………..(iii)

⇒ $I = π \underset{0}{\int^{\frac{π}{2}}} \log \sin (\frac{π}{2} - x) d x$

⇒ $I = π \underset{0}{\int^{\frac{π}{2}}} \log \cos x d x$ ……………(iv)

Adding eq (iii) and (iv), we get

⇒ $2 I = π \underset{0}{\int^{\frac{π}{2}}} (\log \sin x + \log \cos x) d x$

⇒ $2 I = π \underset{0}{\int^{\frac{π}{2}}} (\log \sin x . \cos x) d x$

⇒ $2 I = π \underset{0}{\int^{\frac{π}{2}}} (\log \frac{2}{2} \sin x . \cos x) d x$

⇒ $2 I = π \underset{0}{\int^{\frac{π}{2}}} (\log \frac{\sin 2 x}{2}) d x$

⇒ $2 I = π \underset{0}{\int^{\frac{π}{2}}} (\log \sin 2 x - \log 2) d x$

⇒ $2 I = π \underset{0}{\int^{\frac{π}{2}}} \log \sin 2 x d x - π \underset{0}{\int^{\frac{π}{2}}} \log 2 d x$

⇒ $2 I = π \underset{0}{\int^{\frac{π}{2}}} \log \sin 2 x d x - π \log 2 \times \frac{π}{2}$

Put $2 x = t \Rightarrow d x = \frac{d t}{2}$

Here, at $x = 0 \to t = 0$ and at $x = \frac{π}{2} \to t = π$

⇒ $2 I = \frac{π}{2} \underset{0}{\int^{π}} \log \sin t d t - \frac{π^{2}}{2} \log 2$

⇒ $2 I = \frac{π}{2} \times 2 \underset{0}{\int^{\frac{π}{2}}} \log \sin t d t - \frac{π^{2}}{2} \log 2$

⇒ $2 I = π \underset{0}{\int^{\frac{π}{2}}} \log \sin t d t - \frac{π^{2}}{2} \log 2$

⇒ $2 I = I - \frac{π^{2}}{2} \log 2$ [from eq (iii)]

⇒ $I = - \frac{π^{2}}{2} \log 2$

Hence, $\underset{0}{\int^{π}} x \log \sin x d x = - \frac{π^{2}}{2} \log 2.$

47. $\underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log (\sin x + \cos x) d x$

Ans: Let $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log (\sin x + \cos x) d x$ ……….. (i)

⇒ $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log [\sin (\frac{π}{4} - \frac{π}{4} - x) + \cos (\frac{π}{4} - \frac{π}{4} - x)] d x$

⇒ $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log [\sin (- x) + \cos (- x)] d x$

⇒ $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log (- \sin x + \cos x) d x$

⇒ $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log (\cos x - \sin x) d x$ …………..(ii)

Adding eq (i) and (ii),we get

⇒ 2 $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log (\sin x + \cos x) d x + \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log (\cos x - \sin x) d x$

⇒ 2 $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log [(\sin x + \cos x) . (\cos x - \sin x)] d x$

⇒ 2 $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log [c o s^{2} x - s i n^{2} x] d x$

⇒ 2 $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log \cos 2 x d x$

⇒ 2 $I = 2 \underset{0}{\int^{\frac{π}{4}}} \log \cos 2 x d x$

⇒ $I = \underset{0}{\int^{\frac{π}{4}}} \log \cos 2 x d x$ ……………(iii)

Now, put $2 x = t \Rightarrow d x = \frac{d t}{2}$

Here, at $x = 0 \to t = 0$ and at $x = \frac{π}{4} \to t = \frac{π}{2}$

Therefore,

⇒ $I = \frac{1}{2} \underset{0}{\int^{\frac{π}{2}}} \log \cos t d t$ ………… (iv)

⇒ $I = \frac{1}{2} \underset{0}{\int^{\frac{π}{2}}} \log \cos (\frac{π}{2} - t) d t$

⇒ $I = \frac{1}{2} \underset{0}{\int^{\frac{π}{2}}} \log \sin t d t$ …………..(v)

Adding eq (iv) and (v), we get

⇒ $2 I = \frac{1}{2} \underset{0}{\int^{\frac{π}{2}}} \log \cos t d t + \frac{1}{2} \underset{0}{\int^{\frac{π}{2}}} \log \sin t d t$

⇒ $2 I = \frac{1}{2} \underset{0}{\int^{\frac{π}{2}}} log sin t \cos t d t$

⇒ $4 I = \underset{0}{\int^{\frac{π}{2}}} \log \frac{2}{2} \sin t \cos t d t$

⇒ $4 I = \underset{0}{\int^{\frac{π}{2}}} \log \frac{\sin 2 t}{2} d t$

⇒ $4 I = \underset{0}{\int^{\frac{π}{2}}} (\log \sin 2 t - \log 2) d t$

⇒ $4 I = \underset{0}{\int^{\frac{π}{2}}} (\log \sin (\frac{π}{2} - 2 t)) d t - \underset{0}{\int^{\frac{π}{2}}} \log 2 d t$

⇒ $4 I = \underset{0}{\int^{\frac{π}{2}}} (\log \cos 2 t) d t - \frac{π}{2} \log 2$

⇒ $4 I = 2 \underset{0}{\int^{\frac{π}{4}}} (\log \cos 2 t) d t - \frac{π}{2} \log 2$

⇒ $4 I = 2 I - \frac{π}{2} \log 2$

⇒ $2 I = - \frac{π}{2} \log 2$

⇒ $I = - \frac{π}{4} \log 2$

Hence, $\underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \log (\sin x + \cos x) d x = - \frac{π}{4} \log 2.$

Objective Type Questions

Choose the correct option from given four options in each of the Exercises from 48 to 63.

48. $\int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$ is equal to

(A) $2 (\sin x + x \cos θ) + C$

(B) $2 (\sin x - x \cos θ) + C$

(D) $2 (\sin x - 2 x \cos θ) + C$

Ans: Let $I = \int \frac{\cos 2 x - \cos 2 θ}{\cos x - \cos θ} d x$

⇒ $I = \int \frac{2 c o s^{2} x - 1 - (2 c o s^{2} θ - 1)}{\cos x - \cos θ} d x$

⇒ $I = \int \frac{2 c o s^{2} x - 1 - 2 c o s^{2} θ + 1}{\cos x - \cos θ} d x$

⇒ $I = \int \frac{2 c o s^{2} x - 2 c o s^{2} θ}{\cos x - \cos θ} d x$

⇒ $I = \int \frac{2 (c o s^{2} x - c o s^{2} θ)}{(\cos x - \cos θ)} d x$

⇒ $I = 2 \int (\cos x + \cos θ) d x$

⇒ $I = 2 (\sin x + x \cos θ) + C$

Hence, option (A) is correct answer.

49. $\int \frac{d x}{\sin (x - a) \sin (x - b)}$ is equal to

(A) $\sin (b - a) \log | \frac{\sin (x - b)}{\sin (x - a)} | + C$

(B) $cosec (b - a) \log | \frac{\sin (x - a)}{\sin (x - b)} | + C$

(D) $\sin (b - a) \log | \frac{\sin (x - b)}{\sin (x - a)} | + C$

Ans: Let $I = \int \frac{d x}{\sin (x - a) \sin (x - b)}$

⇒ $I = \frac{1}{\sin (b - a)} \int \frac{\sin (b - a) d x}{\sin (x - a) \sin (x - b)}$

⇒ $I = \frac{1}{\sin (b - a)} \int \frac{\sin (x - a - x + b)}{\sin (x - a) \sin (x - b)} d x$

⇒ $I = \frac{1}{\sin (b - a)} \int \frac{\sin {(x - a) - (x - b)}}{\sin (x - a) \sin (x - b)} d x$

⇒ $I = \frac{1}{\sin (b - a)} \int \frac{\sin (x - a) . \cos (x - b) - \sin (x - b) . \cos (x - a)}{\sin (x - a) \sin (x - b)} d x$

⇒ $I = \frac{1}{\sin (b - a)} \int \frac{\cos (x - b)}{\sin (x - b)} d x - \frac{1}{\sin (b - a)} \int \frac{\cos (x - a)}{\sin (x - a)} d x$

⇒ $I = \frac{1}{\sin (b - a)} \int \cot (x - b) d x - \frac{1}{\sin (b - a)} \int \cot (x - a) d x$

⇒ $I = \frac{1}{\sin (b - a)} \log | \sin (x - b) | - \frac{1}{\sin (b - a)} \log | \sin (x - a) | + C$

⇒ $I = \frac{1}{\sin (b - a)} [\log | \sin (x - b) | - \log | \sin (x - a) |] + C$

⇒ $I = \frac{1}{\sin (b - a)} \log | \frac{(x - b)}{(x - a)} | + C$

⇒ $I = cosec (b - a) \log | \frac{(x - b)}{(x - a)} | + C$

Hence, option (C) is correct answer.

50. $\int t a n^{- 1} \sqrt{x} d x$ is equal to

(A) $(x + 1) t a n^{- 1} \sqrt{x} - \sqrt{x} + C$

(B) $x t a n^{- 1} \sqrt{x} - \sqrt{x} + C$

(D) $\sqrt{x} - (x + 1) t a n^{- 1} \sqrt{x} + C$

Ans: Let $I = \int t a n^{- 1} \sqrt{x} d x$

⇒ $I = \int 1. t a n^{- 1} \sqrt{x} d x$

⇒ $I = t a n^{- 1} \sqrt{x} \int 1 d x - \int (\frac{d}{d x} t a n^{- 1} \sqrt{x} . \int 1 d x) d x$

⇒ $I = x t a n^{- 1} \sqrt{x} - \int (\frac{1}{1 + x} . \frac{1}{2 \sqrt{x}} . x) d x$

⇒ $I = x t a n^{- 1} \sqrt{x} - \frac{1}{2} \int (\frac{\sqrt{x}}{1 + x}) d x$

Now, put $x = t^{2}$

⇒ $d x = 2 t d t$

Therefore,

⇒ $I = x t a n^{- 1} \sqrt{x} - \frac{1}{2} \int \frac{t}{1 + t^{2}} . 2 t d t$

⇒ $I = x t a n^{- 1} \sqrt{x} - \int \frac{t^{2}}{1 + t^{2}} d t$

⇒ $I = x t a n^{- 1} \sqrt{x} - \int \frac{1 + t^{2} - 1}{1 + t^{2}} d t$

⇒ $I = x t a n^{- 1} \sqrt{x} - \int (1 - \frac{1}{1 + t^{2}}) d t$

⇒ $I = x t a n^{- 1} \sqrt{x} - \int 1 d t + \int (\frac{1}{1 + t^{2}}) d t$

⇒ $I = x t a n^{- 1} \sqrt{x} - t + t a n^{- 1} t + C$

⇒ $I = x t a n^{- 1} \sqrt{x} - \sqrt{x} + t a n^{- 1} \sqrt{x} + C$

⇒ $I = (x + 1) t a n^{- 1} \sqrt{x} - \sqrt{x} + C$

Hence, option (A) is correct answer.

51. $\int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x$ is equal to

(A) $\frac{e^{x}}{1 + x^{2}} + C$

(B) $\frac{- e^{x}}{1 + x^{2}} + C$

(D) $\frac{- e^{x}}{{(1 + x^{2})}^{2}} + C$

Ans: Let $I = \int e^{x} {(\frac{1 - x}{1 + x^{2}})}^{2} d x$

⇒ $I = \int e^{x} [\frac{1 + x^{2} - 2 x}{{(1 + x^{2})}^{2}}] d x$

⇒ $I = \int e^{x} [\frac{1 + x^{2}}{{(1 + x^{2})}^{2}} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x$

⇒ $I = \int e^{x} [\frac{1}{(1 + x^{2})} - \frac{2 x}{{(1 + x^{2})}^{2}}] d x$

⇒ $I = \frac{e^{x}}{(1 + x^{2})} + C$

Hence, option (A) is correct answer.

52. $\int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$ is equal to

(A) $\frac{1}{5 x} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

(B) $\frac{1}{5} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

(D) $\frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

Ans: Let $I = \int \frac{x^{9}}{{(4 x^{2} + 1)}^{6}} d x$

⇒ $I = \int \frac{x^{9}}{{[x^{2} (4 + \frac{1}{x^{2}})]}^{6}} d x$

⇒ $I = \int \frac{x^{9}}{x^{12} {(4 + \frac{1}{x^{2}})}^{6}} d x$

⇒ $I = \int \frac{d x}{x^{3} {(4 + \frac{1}{x^{2}})}^{6}}$

Now, put $4 + \frac{1}{x^{2}} = t$

⇒ $- \frac{2}{x^{3}} d x = d t$

⇒ $\frac{d x}{x^{3}} = - \frac{d t}{2}$

Therefore,

⇒ $I = - \frac{1}{2} \int \frac{d t}{{(t)}^{6}}$

⇒ $I = - \frac{1}{2} \int t^{- 6} d t$

⇒ $I = - \frac{1}{2} \times \frac{t^{- 5}}{- 5} + C$

⇒ $I = \frac{1}{10} {(4 + \frac{1}{x^{2}})}^{- 5} + C$

Hence, option (D) is correct answer.

53. $\int \frac{d x}{(x + 2) (x^{2} + 1)} = a \log | 1 + x^{2} | + b t a n^{- 1} x + \frac{1}{5} \log | x + 2 | + C,$ then

(A). $a = - \frac{1}{10}, b = - \frac{2}{5}$

(B) $a = \frac{1}{10}, b = - \frac{2}{5}$

(D) $a = \frac{1}{10}, b = \frac{2}{5}$

Ans: Let $I = \int \frac{d x}{(x + 2) (x^{2} + 1)}$

Now, $\frac{1}{(x + 2) (x^{2} + 1)} = \frac{A}{(x + 2)} + \frac{B x + C}{(x^{2} + 1)}$

⇒ $\frac{1}{(x + 2) (x^{2} + 1)} = \frac{A (x^{2} + 1) + (B x + C) (x + 2)}{(x^{2} + 1)}$

⇒ $1 = A (x^{2} + 1) + (B x + C) (x + 2)$

⇒ $1 = A x^{2} + A + B x^{2} + 2 B x + C x + 2 C$

⇒ $1 = x^{2} (A + B) + x (2 B + C) + (A + 2 C)$

On comparing both sides, we get

⇒ $A + B = 0, 2 B + C = 0$ and $A + 2 C = 1$

From above equations, we get

⇒ $A = \frac{1}{5}, B = - \frac{1}{5}$ and $C = \frac{2}{5}$

Therefore, $\frac{1}{(x + 2) (x^{2} + 1)} = \frac{1}{5 (x + 2)} + \frac{- \frac{x}{5} + \frac{2}{5}}{(x^{2} + 1)}$

⇒ $\int \frac{d x}{(x + 2) (x^{2} + 1)} = \int \frac{d x}{5 (x + 2)} + \int \frac{- \frac{x}{5} + \frac{2}{5}}{(x^{2} + 1)} d x$

⇒ $\int \frac{d x}{(x + 2) (x^{2} + 1)} = \int \frac{d x}{5 (x + 2)} - \frac{1}{5} \int \frac{x}{(x^{2} + 1)} d x + \frac{2}{5} \int \frac{1}{(x^{2} + 1)} d x$

$= \frac{1}{5} \log | x + 5 | - \frac{1}{10} \log | 1 + x^{2} | + \frac{2}{5} t a n^{- 1} x + C$

Here, $a = - \frac{1}{10}$ and $b = \frac{2}{5}$

Hence, option (C) is correct answer.

54. $\int \frac{x^{3}}{x + 1} d x$ is equal to

(A) $x + \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log | 1 - x | + C$

(B) $x + \frac{x^{2}}{2} - \frac{x^{3}}{3} - \log | 1 - x | + C$

(D) $x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \log | 1 + x | + C$

Ans: Let $I = \int \frac{x^{3}}{x + 1} d x$

⇒ $I = \int \frac{x^{3} + 1 - 1}{x + 1} d x$

⇒ $I = \int \frac{x^{3} + 1}{x + 1} d x - \int \frac{1}{x + 1} d x$

⇒ $I = \int (x^{2} - x + 1) d x - \int \frac{1}{x + 1} d x$

⇒ $I = \frac{x^{3}}{3} - \frac{x^{2}}{2} + x - \log | x + 1 | + C$

Hence, option (D) is correct answer.

55. $\int \frac{x + \sin x}{1 + \cos x} d x$ is equal to

(A) $\log | 1 + \cos x | + C$

(B) $\log | x + \sin x | + C$

(D) $x \tan \frac{x}{2} + C$

Ans: Let $I = \int \frac{x + \sin x}{1 + \cos x} d x$

⇒ $I = \int \frac{x}{1 + \cos x} d x + \int \frac{\sin x}{1 + \cos x} d x$

⇒ $I = I_{1} + I_{2}$ ………(i), where, $I_{1} = \int \frac{x}{1 + \cos x} d x$ & $I_{2} = \int \frac{\sin x}{1 + \cos x} d x$

Now, $I_{1} = \int \frac{x}{1 + \cos x} d x$

⇒ $I_{1} = \int \frac{x}{2 c o s^{2} \frac{x}{2}} d x$

⇒ $I_{1} = \frac{1}{2} \int x s e c^{2} \frac{x}{2} d x$

⇒ $I_{1} = \frac{1}{2} [x \int s e c^{2} \frac{x}{2} d x - \int (\frac{d}{d x} x . \int s e c^{2} \frac{x}{2} d x) d x]$

⇒ $I_{1} = \frac{1}{2} [2 x \tan \frac{x}{2} - 2 \int \tan \frac{x}{2} d x] + C$

⇒ $I_{1} = x \tan \frac{x}{2} - \int \tan \frac{x}{2} d x + C$ …………. (ii)

And also we have,

⇒ $I_{2} = \int \frac{\sin x}{1 + \cos x} d x$

⇒ $I_{2} = \int \frac{2 \sin \frac{x}{2} . \cos \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} d x$

⇒ $I_{2} = \int \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} d x$

⇒ $I_{2} = \int \tan \frac{x}{2} d x$ …………..(iii)

Now, put the value of $I_{1}$ and $I_{2}$ in eq (i),

⇒ $I = x \tan \frac{x}{2} - \int \tan \frac{x}{2} d x + C + \int \tan \frac{x}{2} d x$

⇒ $I = x \tan \frac{x}{2} + C$

Hence, option (D) is correct answer.

56. $\int \frac{x^{3} d x}{\sqrt{1 + x^{2}}} = a {(1 + x^{2})}^{\frac{3}{2}} + b \sqrt{1 + x^{2}} + C,$ then

(A) $a = \frac{1}{3}, b = 1$

(B) $a = \frac{- 1}{3}, b = 1$

(D) $a = \frac{1}{3}, b = - 1$

Ans: Let $I = \int \frac{x^{3} d x}{\sqrt{1 + x^{2}}}$

⇒ $I = \int \frac{x^{2} . x d x}{\sqrt{1 + x^{2}}}$

Put $1 + x^{2} = t^{2}$

⇒ $2 x d x = 2 t d t$

⇒ $x d x = t d t$

Therefore,

⇒ $I = \int \frac{(t^{2} - 1) t d t}{t}$

⇒ $I = \int (t^{2} - 1) d t$

⇒ $I = \frac{t^{3}}{3} - t + C$

⇒ $I = \frac{{(\sqrt{1 + x^{2}})}^{3}}{3} - \sqrt{1 + x^{2}} + C$

⇒ $I = \frac{{(1 + x^{2})}^{\frac{3}{2}}}{3} - \sqrt{1 + x^{2}} + C$

Here, $a = \frac{1}{3}, b = - 1$

Hence, option (D) is correct answer.

57 $\underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \frac{d x}{1 + \cos 2 x}$ is equal to

(A) 1

(B) 2

(D) 4

Ans: Let $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \frac{d x}{1 + \cos 2 x}$

⇒ $I = \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} \frac{d x}{2 c o s^{2} x}$

⇒ $I = \frac{1}{2} \underset{- \frac{π}{4}}{\int^{\frac{π}{4}}} s e c^{2} x d x$

⇒ $I = \frac{1}{2} {[\tan x]}_{- \frac{π}{4}}^{\frac{π}{4}}$

⇒ $I = \frac{1}{2} [\tan \frac{π}{4} - \tan (- \frac{π}{4})]$

⇒ $I = \frac{1}{2} [1 + 1]$

⇒ $I = 1$

Hence, option (A) is correct answer.

58. $\underset{0}{\int^{\frac{π}{2}}} \sqrt{1 - \sin 2 x} d x$ is equal to

(A) $2 \sqrt{2}$

(B) $2 (\sqrt{2} + 1)$

(D) $2 (\sqrt{2} - 1)$

Ans: Let $I = \underset{0}{\int^{\frac{π}{2}}} \sqrt{1 - \sin 2 x} d x$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \sqrt{s i n^{2} x + c o s^{2} x - 2 \sin x . \cos x} d x$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} \sqrt{{(\sin x - \cos x)}^{2}} d x$

⇒ $I = \underset{0}{\int^{\frac{π}{2}}} | \sin x - \cos x | d x$

Since, $(\sin x - \cos x) \leq 0$ , $x \in [0, \frac{π}{4}]$ and $(\sin x - \cos x) \geq 0$ , $x \in [\frac{π}{4}, \frac{π}{2}]$

Therefore,

⇒ $I = - \underset{0}{\int^{\frac{π}{4}}} (\sin x - \cos x) d x + \underset{\frac{π}{4}}{\int^{\frac{π}{2}}} (\sin x - \cos x) d x$

⇒ $I = - {[- \cos x - \sin x]}_{0}^{\frac{π}{4}} + {[- \cos x - \sin x]}_{\frac{π}{4}}^{\frac{π}{2}}$

⇒ $I = {[\cos x + \sin x]}_{0}^{\frac{π}{4}} - {[\cos x + \sin x]}_{\frac{π}{4}}^{\frac{π}{2}}$

⇒ $I = [\cos \frac{π}{4} + \sin \frac{π}{4} - 1 - 0] - [\cos \frac{π}{2} + \sin \frac{π}{2} - \cos \frac{π}{4} - \sin \frac{π}{4}]$

⇒ $I = [\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - 1] - [1 - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}]$

⇒ $I = [\sqrt{2} - 1] - [1 - \sqrt{2}]$

⇒ $I = \sqrt{2} - 1 - 1 + \sqrt{2}$

⇒ $I = 2 \sqrt{2} - 2$

⇒ $I = 2 (\sqrt{2} - 1)$

Hence, option (D) is correct answer.

59. $\underset{0}{\int^{\frac{π}{2}}} \cos x e^{\sin x} d x$ is equal to …………..

Ans: Let $I = \underset{0}{\int^{\frac{π}{2}}} \cos x e^{\sin x} d x$

Put $\sin x = t$

⇒ $\cos x d x = d t$

Now, at $x = 0 \Rightarrow t = 0$ and at $x = \frac{π}{2} \Rightarrow t = 1$

Therefore,

⇒ $I = \underset{0}{\int^{1}} e^{t} d x$

⇒ $I = {[e^{t}]}_{0}^{1}$

⇒ $I = e^{1} - e^{0}$

⇒ $I = e - 1$

Hence, $\underset{0}{\int^{\frac{π}{2}}} \cos x e^{\sin x} d x = e - 1.$

60. $\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x$ = ………

Ans: Let $I = \int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x$

⇒ $I = \int e^{x} [\frac{x + 4 - 1}{{(x + 4)}^{2}}] d x$

⇒ $I = \int e^{x} [\frac{x + 4}{{(x + 4)}^{2}} - \frac{1}{{(x + 4)}^{2}}] d x$

⇒ $I = \int e^{x} [\frac{1}{x + 4} - \frac{1}{{(x + 4)}^{2}}] d x$

⇒ $I = \frac{e^{x}}{x + 4} + C$

Hence, $\int \frac{x + 3}{{(x + 4)}^{2}} e^{x} d x = \frac{e^{x}}{x + 4} + C .$

Fill in the blanks in each of the following Exercise 60 to 63.

61. $\underset{0}{\int^{a}} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8}$ , then $a =$ ………..

Ans: here, we have $\underset{0}{\int^{a}} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8}$

⇒ $\frac{1}{4} \underset{0}{\int^{a}} \frac{1}{\frac{1}{4} + x^{2}} d x = \frac{π}{8}$

⇒ $\underset{0}{\int^{a}} \frac{1}{{(\frac{1}{2})}^{2} + x^{2}} d x = \frac{π}{2}$

⇒ ${[\frac{1}{\frac{1}{2}} t a n^{- 1} (\frac{x}{\frac{1}{2}})]}_{0}^{a} = \frac{π}{2}$

⇒ ${[2 t a n^{- 1} (2 x)]}_{0}^{a} = \frac{π}{2}$

⇒ $t a n^{- 1} (2 a) - t a n^{- 1} (0) = \frac{π}{4}$

⇒ $t a n^{- 1} (2 a) = \frac{π}{4}$

⇒ $2 a = \tan \frac{π}{4}$

⇒ $2 a = 1$

⇒ $a = \frac{1}{2}$

Hence, value of a is $\frac{1}{2}$ .

Therefore, $\underset{0}{\int^{a}} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8}$ , then $a = \frac{1}{2} .$

62. $\int \frac{\sin x}{3 + 4 c o s^{2} x} d x =$

Ans: Let $I = \int \frac{\sin x}{3 + 4 c o s^{2} x} d x$

Put $\cos x = t$

⇒ $- \sin x d x = d t$

⇒ $\sin x d x = - d t$

Therefore,

⇒ $I = - \int \frac{d t}{3 + 4 t^{2}}$

⇒ $I = - \frac{1}{4} \int \frac{d t}{\frac{3}{4} + t^{2}}$

⇒ $I = - \frac{1}{4} \int \frac{d t}{{(\frac{\sqrt{3}}{2})}^{2} + t^{2}}$

⇒ $I = - \frac{1}{4} . \frac{1}{\frac{\sqrt{3}}{2}} t a n^{- 1} (\frac{t}{\frac{\sqrt{3}}{2}}) + C$

⇒ $I = - \frac{1}{2 \sqrt{3}} t a n^{- 1} (\frac{2 t}{\sqrt{3}}) + C$

⇒ $I = - \frac{1}{2 \sqrt{3}} t a n^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C$

Hence, $\int \frac{\sin x}{3 + 4 c o s^{2} x} d x = - \frac{1}{2 \sqrt{3}} t a n^{- 1} (\frac{2 \cos x}{\sqrt{3}}) + C .$

63. The value of $\int_{- π}^{π} s i n^{3} x . c o s^{2} x d x$ is …………..

Ans: Let $I = \int_{- π}^{π} s i n^{3} x . c o s^{2} x d x$ …………. (i)

⇒ $I = \int_{- π}^{π} s i n^{3} (π - π - x) . c o s^{2} (π - π - x) d x$

⇒ $I = \int_{- π}^{π} s i n^{3} (- x) . c o s^{2} (- x) d x$

⇒ $I = - \int_{- π}^{π} s i n^{3} x . c o s^{2} x d x$ …………….(ii)

Now, adding eq (i) and (ii), we get

⇒ $2 I = 0$

⇒ $I = 0$

Hence, the value of $\int_{- π}^{π} s i n^{3} x . c o s^{2} x d x$ is 0.

Importance of Integrals

Students will get a brief knowledge about the concepts that are mentioned in the PDF. The Exemplar PDF of Chapter 7 - Integrals is designed as per the latest syllabus pattern advised by the Central Board of Secondary Education. The PDF is structured by the expert faculty of Vedantu, specifically structured to enhance the learning and understanding capacity of the students. The students following the Exemplar with full dedication will surely achieve good results.

Class 12 is the most important academic year for students. All the Chapters in all the subjects are very important, students do have pressure to perform well in the Exams. Chapter 7 Integrals is such an important Chapter that focuses on the concepts used in other Chapters as well. This Chapter increases the thinking process of the students, questions asked from this Chapter are not difficult so this Chapter can be quite scoring for all students.

Students neglecting the concepts of this Chapter face trouble as the concepts present in this Chapter also help to solve other Chapter questions as well. NCERT Exemplar of Chapter 7 consists of a textbook solution for Class 12. There are many important concepts but the most important concepts are mentioned below. This Chapter basically focuses on the questions asked on topics like -

Integrations
Properties of Integrals
Geometrical representation
Comparison between integration and differentiation
Definite Integral - limit of the sum
Different
methods of integration
Properties of definite Integral and some other important question

Those students who want to have a good grip on the concepts must focus on the structured solutions mentioned in the Exemplar. Students can register themselves on Vedantu’s website for other important topics. Vedantu also offers online study support to the students, the online present material consists of additional questions, notes, sample papers, Exemplars. All the study material is designed and approved by the community of faculty that consists of a group of teachers that focuses on the quality of the content present from the Vedantu’s end.

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FAQs on NCERT Exemplar for Class 12 Maths Chapter-7 (Book Solutions)

1. From where do students can get the Exemplar for Chapter 7 Integral for Class 12?

The NCERT Exemplar for Chapter 7 Integral of Class 12 is available on Vedantu.com. Students reach out to the website and can get a free PDF for Chapter 7. The PDF of Chapter 7 - Integrals consists of important topics and additional questions that will help students to learn and practice to score more marks in their Examinations. The PDF is available at Vedantu’s official website for free.

2. Is Chapter 7 important for Class 12 students?

Class 12 is the most important Class for students as well as the toughest Class. Students do have mental pressure to perform well in their boards. All the chapters of every subject are important as Chapter 7 Integrals. This Chapter increases the thinking process of the students, questions asked from this Chapter are not really tough though this Chapter can be a score booster for all students. Students neglecting the concepts of this Chapter go through problems as the concepts present in the respective Chapter help to solve other Chapter questions as well.

3. What are the important concepts for the Chapter 7 Integrals?

Chapter 7 Integrals do have many important concepts and the most important topics are -Integrations, Properties of Integrals, Geometrical representation, Comparison between integration and differentiation , Definite Integral - limit of the sum, Different methods of integration, Properties of definite Integral, and some other important question. Students focusing on these topics will surely achieve good marks.

4. How is Vedantu beneficial for Class 12 students?

Vedantu is a website that focuses on the development of students. This site helps students to avail important study material like notes, sample papers, Exemplars, additional questions, and others. Students can get free study by just registering themselves on Vedantu’s official website i.e. Vedantu.com. Vedantu also offers live Classes for the students to re-understand the concepts taught in Class. Students can clear their doubts through live sessions.

NCERT Exemplar for Class 12 Maths Chapter-7 (Book Solutions)

Free PDF Download for NCERT Exemplar Available

Access NCERT Exemplar Solutions for CBSE Class 12 Mathematics Chapter 7 - Integrals

Solved Examples

Short Answer (S.A.)

Long Answer (L.A.)

Objective Type Questions

Choose the correct answer from the given four option in each of the Examples from 20 to 30.

Fill in the blanks in each of the 29 to 32.

Exercise 7.3

Short Answer (S.A)

Verify the following:

Evaluate the Following:

Evaluate the Following As Limit of Sums:

Evaluate the Following:

Long Answer (L.A)

Objective Type Questions

Choose the correct option from given four options in each of the Exercises from 48 to 63.

Fill in the blanks in each of the following Exercise 60 to 63.

Importance of Integrals

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