NCERT Solutions for Class 12 Maths Chapter 7: Integrals - Exercise 7.11

Exercise 7.11

1.  Integrate\[\int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} \]

Ans: Let \[I = \int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx} \]

Then, by the property \[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}\left( {\dfrac{\pi }{2} - x} \right)dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}xdx} \]

Adding two values of \[I\] we get

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {{{\cos }^2}x + {{\sin }^2}x} \right)dx} \]

Solving further we get,

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {1dx} \]

\[2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]

Substitute the limits.

\[2I = \dfrac{\pi }{2}\]

Divide by 2 on both sides.

\[I = \dfrac{\pi }{4}\]

Therefore,

\[\int\limits_0^{\dfrac{\pi }{2}} {{{\cos }^2}xdx}  = \dfrac{\pi }{4}\]

2. Integrate \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x}  + \sqrt {\cos x} }}dx} \]

Ans: \[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x}  + \sqrt {\cos x} }}dx} \]

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin \left( {\dfrac{\pi }{2} - x} \right)} }}{{\sqrt {\sin \left( {\dfrac{\pi }{2} - x} \right)}  + \sqrt {\cos \left( {\dfrac{\pi }{2} - x} \right)} }}dx} \]

Solving further we get

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\cos x}  + \sqrt {\sin x} }}dx} \]

Adding two values of \[I\] we get

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x}  + \sqrt {\cos x} }}{{\sqrt {\cos x}  + \sqrt {\sin x} }}dx} \]

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {1dx} \]

\[2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]

Substitute the limits.

\[2I = \dfrac{\pi }{2}\]

Divide by 2 on both sides.

\[I = \dfrac{\pi }{4}\]

3. Integrate  \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{3}{2}}}x}}{{{{\sin }^{\dfrac{3}{2}}}x + {{\cos }^{\dfrac{3}{2}}}x}}dx} \]

Ans: Let \[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{3}{2}}}x}}{{{{\sin }^{\dfrac{3}{2}}}x + {{\cos }^{\dfrac{3}{2}}}x}}dx} \]

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{3}{2}}}\left( {\dfrac{\pi }{2} - x} \right)}}{{{{\sin }^{\dfrac{3}{2}}}\left( {\dfrac{\pi }{2} - x} \right) + {{\cos }^{\dfrac{3}{2}}}\left( {\dfrac{\pi }{2} - x} \right)}}dx} \]

Simplifying further we get,

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^{\dfrac{3}{2}}}x}}{{{{\sin }^{\dfrac{3}{2}}}x + {{\cos }^{\dfrac{3}{2}}}x}}dx} \]

Adding two values of \[I\] we get

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^{\dfrac{3}{2}}}x + {{\cos }^{\dfrac{3}{2}}}x}}{{{{\sin }^{\dfrac{3}{2}}}x + {{\cos }^{\dfrac{3}{2}}}x}}dx} \]

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {1dx} \]

\[2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]

Substitute the limits.

\[2I = \dfrac{\pi }{2}\]

Divide on both sides by 2.

\[I = \dfrac{\pi }{4}\]

4. Integrate \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}dx} \]

Ans: \[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}dx} \]

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\cos }^5}\left( {\dfrac{\pi }{2} - x} \right)}}{{{{\sin }^5}\left( {\dfrac{\pi }{2} - x} \right) + {{\cos }^5}\left( {\dfrac{\pi }{2} - x} \right)}}dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}dx} \]

Adding two values of \[I\] we get

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{{{\sin }^5}x + {{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}dx} \]

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {1dx} \]

\[2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]

\[2I = \dfrac{\pi }{2}\]

Divide by 2 on both sides.

\[I = \dfrac{\pi }{4}\]

5. Integrate  \[\int\limits_{ - 5}^5 {|x + 2|dx} \]

Ans: Let \[I = \int\limits_{ - 5}^5 {|x + 2|dx} \]

We know that \[\left( {x + 2} \right) \leqslant 0\]

\[\left[ { - 5, - 2} \right]\] and \[\left( {x + 2} \right) \geqslant 0\] on \[\left[ { - 2,5} \right]\].

Then by the property, 

\[\int\limits_a^b {f\left( x \right)dx}  = \int\limits_a^c {f\left( x \right)dx}  + \int\limits_c^b {f\left( x \right)dx} \]

\[I = \int\limits_{ - 5}^{ - 2} { - \left( {x + 2} \right)dx}  + \int\limits_{ - 2}^5 {\left( {x + 2} \right)dx} \]

\[I =  - \left[ {\dfrac{{{x^2}}}{2} + 2x} \right]_{ - 5}^{ - 2} + \left[ {\dfrac{{{x^2}}}{2} + 2x} \right]_{ - 2}^5\]

\[ =  - \left[ {\dfrac{{{{\left( { - 2} \right)}^2}}}{2} + 2\left( { - 2} \right) - \dfrac{{{{\left( { - 5} \right)}^2}}}{2} - 2\left( { - 5} \right)} \right] + \left[ {\dfrac{{{{\left( 5 \right)}^2}}}{2} + 2\left( 5 \right) - \dfrac{{{{\left( { - 2} \right)}^2}}}{2} - 2\left( { - 2} \right)} \right]\]

\[ =  - \left[ {2 - 4 - \dfrac{{25}}{2} + 10} \right] + \left[ {\dfrac{{25}}{2} + 10 - 2 + 4} \right]\]

\[ =  - 2 + 4 + \dfrac{{25}}{2} - 10 + \dfrac{{25}}{2} + 10 - 2 + 4\]

\[ = 29\]

6. Integrate  \[\int\limits_2^8 {|x - 5|dx} \]

Ans: Let \[I = \int\limits_2^8 {|x - 5|dx} \]

We know that   \[\left( {x - 5} \right) \leqslant 0\] on \[\left[ {2,5} \right]\] and \[\left( {x - 5} \right) \geqslant 0\] on \[\left[ {5,8} \right]\].

Then by the property,

\[\int\limits_a^b {f\left( x \right)dx}  = \int\limits_a^c {f\left( x \right)dx}  + \int\limits_c^b {f\left( x \right)dx} \]

\[I = \int\limits_2^5 { - \left( {x - 5} \right)dx}  + \int\limits_5^8 {\left( {x - 5} \right)dx} \]

\[I =  - \left[ {\dfrac{{{x^2}}}{2} - 5x} \right]_2^5 + \left[ {\dfrac{{{x^2}}}{2} - 5x} \right]_5^8\]

\[ =  - \left[ {\dfrac{{{{\left( 5 \right)}^2}}}{2} - 5\left( 5 \right) - \dfrac{{{{\left( 2 \right)}^2}}}{2} + 5\left( 2 \right)} \right] + \left[ {\dfrac{{{{\left( 8 \right)}^2}}}{2} - 5\left( 8 \right) - \dfrac{{{{\left( 5 \right)}^2}}}{2} + 5\left( 5 \right)} \right]\]

\[ =  - \dfrac{{25}}{2} + 25 + 2 - 10 + 32 - 40 - \dfrac{{25}}{2} + 25\]

\[ = 9\]

7. Integrate \[\int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} \]

Ans: Let \[I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} \]

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

Apply the above property

\[I = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - \left( {1 - x} \right)} \right)}^n}dx} \]

\[ = \int\limits_0^1 {\left( {1 - x} \right){{\left( x \right)}^n}dx} \]

\[ = \int\limits_0^1 {\left( {{x^n} - {x^{n + 1}}} \right)dx} \]

\[ = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}} - \dfrac{{{x^{n + 2}}}}{{n + 2}}} \right]_0^1\]

Substitute the values and solve the integration.

\[ = \left[ {\dfrac{1}{{n + 1}} - \dfrac{1}{{n + 2}}} \right]\]

\[ = \dfrac{{\left( {n + 2} \right) - \left( {n + 1} \right)}}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\]

\[ = \dfrac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\]

8. Integrate  \[\int\limits_0^{\dfrac{\pi }{4}} {\log \left( {1 + \tan x} \right)dx} \]

Ans: Let  \[I = \int\limits_0^{\dfrac{\pi }{4}} {\log \left( {1 + \tan x} \right)dx} \]

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

Apply the property in the integration.

\[I = \int\limits_0^{\dfrac{\pi }{4}} {\log \left[ {1 + \tan \left( {\dfrac{\pi }{4} - x} \right)} \right]dx} \]

\[ = \int\limits_0^{\dfrac{\pi }{4}} {\log \left\{ {1 + \dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}}} \right\}dx} \]

\[ = \int\limits_0^{\dfrac{\pi }{4}} {\log \left\{ {1 + \dfrac{{1 - \tan x}}{{1 + \tan x}}} \right\}dx} \]

\[ = \int\limits_0^{\dfrac{\pi }{4}} {\log \dfrac{2}{{1 + \tan x}}dx} \]

\[ = \int\limits_0^{\dfrac{\pi }{4}} {\log 2dx}  - \int\limits_0^{\dfrac{\pi }{4}} {\log \left( {1 + \tan x} \right)dx} \]

From the equation (1)

\[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\log 2dx}  - I\]

\[ \Rightarrow 2I = \left[ {x\log 2} \right]_0^{\dfrac{\pi }{4}}\]

Substitute the limits in the integration and solve.

\[2I = \dfrac{\pi }{4}\log 2\]

\[I = \dfrac{\pi }{8}\log 2\]

9. Integrate  \[\int\limits_0^2 {x\sqrt {2 - x} dx} \]

Ans: Let \[I = \int\limits_0^2 {x\sqrt {2 - x} dx} \]

Integrate the expression.

\[I = \int\limits_0^2 {\left( {2 - x} \right)\sqrt x dx} \]

\[ = \int\limits_0^2 {\left( {2{x^{\dfrac{1}{2}}} - {x^{\dfrac{3}{2}}}} \right)dx} \]

\[ = \left[ {2\left( {\dfrac{{{x^{\dfrac{3}{2}}}}}{{^{\dfrac{3}{2}}}}} \right) - \left( {\dfrac{{{x^{\dfrac{5}{2}}}}}{{^{\dfrac{5}{2}}}}} \right)} \right]_0^2\]

\[ = \left[ {\dfrac{4}{3}{x^{\dfrac{3}{2}}} - \dfrac{2}{5}{x^{\dfrac{5}{2}}}} \right]_0^2\]

Apply the limits and solve.

\[I = \dfrac{{4 \times 2\sqrt 2 }}{3} - \dfrac{{2 \times 4\sqrt 2 }}{5}\]

\[ = \dfrac{{40\sqrt 2  - 24\sqrt 2 }}{{15}}\]

\[ = \dfrac{{16\sqrt 2 }}{{15}}\]

10. Integrate  \[\int\limits_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log \sin 2x} \right)dx} \]

Ans: Let \[I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log \sin 2x} \right)dx} \]

Integrate the expression.

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log \left( {2\sin x\cos x} \right)} \right)dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log \sin x - \log \cos x - \log 2} \right)dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\log \cos x - \log \sin x - \log 2} \right)dx} \]

Then by the property 

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

Apply the property

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\left( {\log \sin x - \log \cos x - \log 2} \right)dx} \]

Adding two values of \[I\] we get

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\left( { - \log 2 - \log 2} \right)dx} \]

\[2I =  - 2\log 2\int\limits_0^{\dfrac{\pi }{2}} {1dx} \]

Integrate and solve further.

\[I =  - \log 2\left[ {\dfrac{\pi }{2}} \right]\]\[\]

\[I = \left( {\log \dfrac{1}{2}} \right)\left[ {\dfrac{\pi }{2}} \right]\]

\[I = \dfrac{\pi }{2}\log \dfrac{1}{2}\]

11. Integrate  \[\int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {{{\sin }^2}xdx} \]

Ans: Let \[I = \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {{{\sin }^2}xdx} \]

We know that 

\[{\sin ^2}\left( { - x} \right) = {\left( {\sin \left( { - x} \right)} \right)^2} = \left( {{{\left( {\sin x} \right)}^2}} \right) = {\sin ^2}x\]

\[{\sin ^2}x\]is an even function. Then for even function

\[\int\limits_{ - a}^a {f\left( x \right)dx = } 2\int\limits_0^a {f\left( x \right)dx} \]

Use the property and integrate further.

\[I = 2\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}xdx} \]

\[ = 2\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{1 - \cos 2x}}{2}dx} \]

\[ = \int\limits_0^{\dfrac{\pi }{2}} {\left( {1 - \cos 2x} \right)dx} \]

\[ = \left[ {x - \dfrac{{\sin 2x}}{2}} \right]_0^{\dfrac{\pi }{2}}\]

Apply the limits and solve.

\[I = \dfrac{\pi }{2}\]

12. Integrate  \[\int\limits_0^\pi  {\dfrac{{xdx}}{{1 + \sin x}}} \]

Ans: Let \[I = \int\limits_0^\pi  {\dfrac{{xdx}}{{1 + \sin x}}} \]

\[I = \int\limits_0^\pi  {\dfrac{{\left( {\pi  - x} \right)dx}}{{1 + \sin \left( {\pi  - x} \right)}}} \]

\[I = \int\limits_0^\pi  {\dfrac{{\left( {\pi  - x} \right)dx}}{{1 + \sin x}}} \]
Adding two values of \[I\] we get

\[2I = \int\limits_0^\pi  {\dfrac{{\pi dx}}{{1 + \sin x}}} \]

Multiply numerator and denominator by $\left( {1 - \sin x} \right)$ .

\[2I = \pi \int\limits_0^\pi  {\dfrac{{\left( {1 - \sin x} \right)}}{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}dx} \]

\[2I = \pi \int\limits_0^\pi  {\dfrac{{1 - \sin x}}{{{{\cos }^2}x}}dx} \]

Integrate the expression.

\[2I = \pi \int\limits_0^\pi  {\left( {{{\sec }^2}x - \tan x\sec x} \right)dx} \]

Apply the limits.

\[2I = \pi \left( 2 \right)\]

\[I = \pi \]

13. Integrate  \[\int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {{{\sin }^7}xdx} \]

Ans: Let \[I = \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {{{\sin }^7}xdx} \]

We know

\[{\sin ^7}\left( { - x} \right) = {\left( {\sin \left( { - x} \right)} \right)^7} = \left( {{{\left( {\sin x} \right)}^7}} \right) = {\sin ^7}x\]

\[{\sin ^7}x\] is an odd function.

 For any odd function,

\[\int\limits_{ - a}^a {f\left( x \right)dx = } 0\]

Hence, apply the property in the integration and solve.

\[I = \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {{{\sin }^7}xdx} \]

\[ = 0\]

14. Integrate  \[\int\limits_0^{2\pi } {{{\cos }^5}xdx} \]

Ans: Let $I = \int\limits_0^{2\pi } {{{\cos }^5}xdx} $

We know that,

${\cos ^5}\left( {2\pi  - x} \right) = {\cos ^5}x$

It is known that

$\int\limits_0^{2a} {f\left( x \right)dx = } 2\int\limits_0^a {f\left( x \right)dx} ,{\text{   if  }}f\left( {2a - x} \right) = f\left( x \right)$

$  {\text{                  =  0,   if  }}f\left( {2a - x} \right) =  - f\left( x \right) $

$I = 2\int\limits_0^\pi  {{{\cos }^5}xdx} $

From the result,

${\cos ^5}\left( {\pi  - x} \right) =  - {\cos ^5}x$

Hence,

$I = 2 \times 0$

$I = 0$

15. Integrate  \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} \]

Ans: Let $I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} $

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

Hence,

$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sin \left( {\dfrac{\pi }{2} - x} \right) - \cos \left( {\dfrac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\dfrac{\pi }{2} - x} \right)\cos \left( {\dfrac{\pi }{2} - x} \right)}}dx} $

$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\cos x - \sin x}}{{1 + \sin x\cos x}}dx} $

Adding two values of \[I\] we get

$2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{0}{{1 + \sin x\cos x}}dx} $

Hence,

$I = 0$

16. Integrate  \[\int\limits_0^\pi  {\log \left( {1 + \cos x} \right)dx} \]

Ans:\[I = \int\limits_0^\pi  {\log \left( {1 + \cos x} \right)dx} \]

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

Hence,

\[I = \int\limits_0^\pi  {\log \left( {1 + \cos \left( {\pi  - x} \right)} \right)dx} \]

\[I = \int\limits_0^\pi  {\log \left( {1 - \cos x} \right)dx} \]

 Adding two values of \[I\] we get

$2I = \int\limits_0^\pi  {\left\{ {\log \left( {1 + \cos x} \right) + \log \left( {1 - \cos x} \right)} \right\}dx}$

$2I = \int\limits_0^\pi  {\left\{ {\log \left( {1 - {{\cos }^2}x} \right)} \right\}dx}$

Simplifying further we get,

\[2I = \int\limits_0^\pi  {\log {{\sin }^2}xdx} \]

Use the property, $\log {a^b} = b\log a$ 

Hence,

\[2I = 2\int\limits_0^\pi  {\log \sin xdx} \]

\[I = \int\limits_0^\pi  {\log \sin xdx} \]

We know that,

\[\sin \left( {\pi  - x} \right) = \sin x\]

Hence,

\[I=2\int\limits_{0}^{\pi /2}{\log \sin \left( \frac{\pi }{2}-x \right)}\text{ }\!\!~\!\!\text{ }dx\]

Adding two values of \[I\] we get

\[2I = 2\int\limits_0^\begin{subarray}{l} 

  \pi  \\ 

  2 

\end{subarray}  {\left( {\log \sin x + \log \cos x} \right)dx} \]

\[I = \int\limits_0^\begin{subarray}{l} 

  \pi  \\ 

  2 

\end{subarray}  {\left( {\log \sin x + \log \cos x + \log 2 - \log 2} \right)dx} \]

Use the property, $\log a + \log b = \log ab$ .

Hence,

\[I = \int\limits_0^\begin{subarray}{l} 

  \pi  \\ 

  2 

\end{subarray}  {\left( {\log 2\sin x\cos x - \log 2} \right)dx} \]

Separating the integrals we get,

\[I = \int\limits_0^\begin{subarray}{l} 

  \pi  \\ 

  2 

\end{subarray}  {\log \sin 2xdx}  - \int\limits_0^\begin{subarray}{l} 

  \pi  \\ 

  2 

\end{subarray}  {\log 2dx} \]

Let us assume,

\[2x = t\]

On differentiating both sides w.r.t. $x$  we get,

\[2dx = dt\]

Also

When \[x = 0\], then \[t = 0\]

When \[x = \dfrac{\pi }{2}\], then \[t = \pi \]

\[\therefore I = \dfrac{1}{2}\int\limits_0^\begin{subarray}{l}  \pi  \\  2 \end{subarray}  {\log \sin tdx}  - \dfrac{\pi }{2}\log 2\]

Solving further we get,

\[I = \dfrac{I}{2} - \dfrac{\pi }{2}\log 2\]

\[\dfrac{I}{2} =  - \dfrac{\pi }{2}\log 2\]

\[I =  - \pi \log 2\]

17. Integrate  \[\int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x  + \sqrt {a - x} }}dx} \]

Ans: Let \[I = \int\limits_0^a {\dfrac{{\sqrt x }}{{\sqrt x  + \sqrt {a - x} }}dx} \]

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

Hence,

\[I = \int\limits_0^a {\dfrac{{\sqrt {a - x} }}{{\sqrt {a - x}  + \sqrt x }}dx} \]

Adding two values of \[I\] we get

\[2I = \int\limits_0^a {\dfrac{{\sqrt x  + \sqrt {a - x} }}{{\sqrt x  + \sqrt {a - x} }}dx} \]

\[2I = \int\limits_0^a {1dx} \]

Integrating and applying limits we get,

\[2I = \left[ x \right]_0^a\]

\[2I = a\]

Dividing by 2 on both sides we get,

\[I = \dfrac{a}{2}\]

18. Integrate  \[\int\limits_0^4 {|x - 1|dx} \]

Ans: Let \[I = \int\limits_0^4 {|x - 1|dx} \]

We know that

\[\left( {x - 1} \right) \leqslant 0\]

\[0 \leqslant x \leqslant 1\]

\[\left( {x - 1} \right) \geqslant 0\]

\[1 \leqslant x \leqslant 4\]

Hence, separate the limits to separate the integration.

\[I = \int\limits_0^1 { - \left( {x - 1} \right)dx}  + \int\limits_1^4 {\left( {x - 1} \right)dx} \]

Integrate the expression.

\[I =  - \left[ {x - \dfrac{{{x^2}}}{2}} \right]_0^1 + \left[ {\dfrac{{{x^2}}}{2} - x} \right]_1^4\]

Apply the limits and simply the expression.

\[I = 1 - \dfrac{1}{2} + \dfrac{{{4^2}}}{2} - 4 - \dfrac{1}{2} + 1\]

\[I = 2 - 1 + \dfrac{{{4^2}}}{2} - 4\]

\[I = 5\]

19. Show that \[\int\limits_0^a {f\left( x \right)g\left( x \right)dx = } 2\int\limits_0^a {f\left( x \right)dx} \], if \[f\]and \[g\]are defined as  \[f\left( x \right) = f\left( {a - x} \right)\] and \[g\left( x \right) + g\left( {a - x} \right) = 4\].

Ans: It is given that \[I = \int\limits_0^a {f\left( x \right)g\left( x \right)dx} \]

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

Hence,

\[I = \int\limits_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} \]

\[I = \int\limits_0^a {f\left( x \right)g\left( {a - x} \right)dx} \]

Adding two values of \[I\] we get,

\[2I = \int\limits_0^a {\left\{ {f\left( x \right)g\left( x \right) + f\left( x \right)g\left( {a - x} \right)} \right\}dx} \]

\[2I = \int\limits_0^a {f\left( x \right)\left\{ {g\left( x \right) + g\left( {a - x} \right)} \right\}dx} \]

Substitute the value of \[g\left( x \right) + g\left( {a - x} \right) = 4\] in the integration.

\[2I = \int\limits_0^a {f\left( x \right) \times 4dx} \]

\[I = 2\int\limits_0^a {f\left( x \right)dx} \]

Hence,

\[\int\limits_0^a {f\left( x \right)g\left( x \right)dx = } 2\int\limits_0^a {f\left( x \right)dx} \]

20. The value of \[\int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {\left( {{x^3} + x\cos x + {{\tan }^5}x + 1} \right)} dx\] is

1. 0

2. 2

3. \[\pi \]

4. 1

Ans: \[I = \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {\left( {{x^3} + x\cos x + {{\tan }^5}x + 1} \right)} dx\]

Separate the integrals to integrate separately.

\[I = \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {{x^3}} dx + \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {x\cos x} dx + \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} {{{\tan }^5}x} dx + \int\limits_{\dfrac{{ - \pi }}{2}}^{\dfrac{\pi }{2}} 1 dx\]

For any odd function,

\[\int\limits_{ - a}^a {f\left( x \right)dx = } 0\]

Hence,

\[I = 0 + 0 + 0 + 2\int\limits_0^{\dfrac{\pi }{2}} 1 dx\]

Integrating further and solving we get,

\[I = 2\left[ x \right]_0^{\dfrac{\pi }{2}}\]

\[ = \dfrac{{2\pi }}{2}\]

\[ = \pi \]

20. The value of \[\int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)dx} \] is

1. 2

2. \[\dfrac{3}{4}\]

3. 0

4. 2

Ans:  \[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)dx} \]

Then, by the property

\[\int\limits_0^a {f\left( x \right)dx}  = \int\limits_0^a {f\left( {a - x} \right)dx} \]

Apply the property in the integration.

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left[ {\dfrac{{4 + 3\sin \left( {\dfrac{\pi }{2} - x} \right)}}{{4 + 3\cos \left( {\dfrac{\pi }{2} - x} \right)}}} \right]dx} \]

\[I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)dx} \]

Adding two values of \[I\] we get

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) + \log \left( {\dfrac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} dx\]

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \left( {\dfrac{{4 + 3\sin x}}{{4 + 3\cos x}} \times \dfrac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} dx\]

\[2I = \int\limits_0^{\dfrac{\pi }{2}} {\log 1} dx\]

\[2I = \int\limits_0^{\dfrac{\pi }{2}} 0 dx\]

\[I = 0\]

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.11

Opting for the NCERT solutions for Ex 7.11 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.11 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 7 Exercise 7.11 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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