# NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.6) Exercise 7.6

## NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.6) Exercise 7.6 Exercise 7.6 is an essential exercise from NCERT Class 12 Maths syllabus. The topics present in Exercise 7.6 Maths Class 12 are mainly integration with primary and crucial methods. Integration methods include partial fraction integration, integration by substitution and parts. Exercise 7.6 Class 12 Maths entirely has problems on integrals under the section of integration by factors. Class 12 NCERT Solutions for Maths question and answer set is divided into part 1 and part 2 with a solution book. NCERT Class 12 Maths has a total of 13 chapters. Here, we have provided the solutions for Exercise 7.6 Maths Class 12. The answers will help you prepare for the final examination. Our expert faculty have solved Exercise 7.6 Class 12 Maths as per the latest CBSE guidelines and syllabus.

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EXERCISE 7.3

Refer to pages 28-41 for exercise 7.3 in the PDF

1. Integrate ${\sin ^2}\left( {2x + 5} \right)$ .

Ans:  Let $y = {\sin ^2}\left( {2x + 5} \right)$

As it is not integrable directly, therefore we will use trigonometry formula which is ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$

Therefore, above question becomes,

$y = \dfrac{{1 - \cos 2\left( {2x + 5} \right)}}{2} \hfill \\ y = \dfrac{{1 - \cos \left( {4x + 10} \right)}}{2} \hfill \\$

Integrating w.r.t. $x$ .

$\int {ydx} = \int {\dfrac{{1 - \cos \left( {4x + 10} \right)}}{2}}\hfill \\ \int {ydx} = \dfrac{1}{2}\int {1dx} - \dfrac{1}{2}\int {\cos \left( {4x + 10} \right)} dx \hfill \\ \int {ydx} = \dfrac{1}{2}x - \dfrac{1}{2}\left( {\dfrac{{\sin \left( {4x + 10} \right)}}{4}} \right) + C \hfill \\ \int {ydx} = \dfrac{1}{2}x - \dfrac{1}{8}\sin \left( {4x + 10} \right) + C \hfill \\$

2. Integrate $\sin 3x \cdot \cos 4x$.

Ans:  Let $y = \sin 3x \cdot \cos 4x$

As it is not integrable directly, therefore we will use trigonometry formula which is

$\sin A\cos B = \dfrac{1}{2}\left\{ {\sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right\}$

Therefore, above equation becomes,

$y = \dfrac{1}{2}\left\{ {\sin \left( {3x + 4x} \right) + \sin \left( {3x - 4x} \right)} \right\} \hfill \\ y = \dfrac{1}{2}\left\{ {\sin \left( {7x} \right) + \sin \left( { - x} \right)} \right\} \hfill \\$

Using $\sin \left( { - x} \right) = - \sin x$

$y = \dfrac{1}{2}\left\{ {\sin \left( {7x} \right) - \sin \left( x \right)} \right\}$

Integrating w.r.t. $x$ .

$\int {ydx} = \dfrac{1}{2}\int {\left\{ {\sin \left( {7x} \right) - \sin \left( x \right)} \right\}} dx \hfill \\ \int {ydx} = \dfrac{1}{2}\int {\sin 7x} dx - \dfrac{1}{2}\int {\sin xdx}\hfill \\ \int {ydx} = \dfrac{1}{2}\left( {\dfrac{{ - \cos 7x}}{7}} \right) - \dfrac{1}{2}\left( { - \cos x} \right) + C \hfill \\ \int {ydx} = - \dfrac{{\cos 7x}}{{14}} + \dfrac{{\cos x}}{2} + C \hfill \\$

Where $C$ is an arbitrary constant.

3. Integrate $\cos 2x\cos 4x\cos 6x$ .

Ans:  Let $y = \cos 2x\cos 4x\cos 6x$

As it is not integrable directly, therefore we will use trigonometry formula which is

$\cos A\cos B = \dfrac{1}{2}\left\{ {\cos \left( {A + B} \right) + \cos \left( {A - B} \right)} \right\}$

Therefore, above equation becomes,

$y = \cos 2x\cos 4x\cos 6x \hfill \\ y = \cos 2x\left[ {\dfrac{1}{2}\left\{ {\cos \left( {4x + 6x} \right) + \cos \left( {4x - 6x} \right)} \right\}} \right] \hfill \\ y = \dfrac{1}{2}\left\{ {\cos 2x\cos 10x + \cos 2x\cos \left( { - 2x} \right)} \right\} \hfill \\$

Using $\cos \left( { - x} \right) = \cos x$

$y = \dfrac{1}{2}\left\{ {\cos 2x\cos 10x + {{\cos }^2}2x} \right\}$

Again applying $\cos A\cos B = \dfrac{1}{2}\left\{ {\cos \left( {A + B} \right) + \cos \left( {A - B} \right)} \right\}$and ${\cos ^2}2x = \dfrac{{1 + \cos 4x}}{2}$ ,

$y = \dfrac{1}{2}\left[\dfrac{1}{2} {\left\{ {\cos \left( {2x + 10x} \right) + \cos \left( {2x - 10x} \right)} \right\} + \left( {\dfrac{{1 + \cos 4x}}{2}} \right)} \right] \hfill \\ y = \dfrac{1}{4}\left( {\cos 12x + \cos \left( { - 8x} \right) + 1 + \cos 4x} \right) \hfill \\ y = \dfrac{1}{4}\left( {\cos 12x + \cos 8x + 1 + \cos 4x} \right) \hfill \\$

Integrating w.r.t. $x$ .

$\int {ydx} = \dfrac{1}{4}\int {\left( {\cos 12x + \cos 8x + 1 + \cos 4x} \right)dx}\hfill \\ \int {ydx} = \dfrac{1}{4}\left[ {\dfrac{{\sin 12x}}{{12}} + \dfrac{{\sin 8x}}{8} + x + \dfrac{{\sin 4x}}{4} + C} \right] \hfill \\ \int {ydx} = \dfrac{{\sin 12x}}{{48}} + \dfrac{{\sin 8x}}{{32}} + \dfrac{x}{4} + \dfrac{{\sin 4x}}{{16}} + C \hfill \\$

Where $C$ is an arbitrary constant.

4. Integrate${\sin ^3}\left( {2x + 1} \right)$

Ans:  Let $I = \int {{{\sin }^3}\left( {2x + 1} \right)} dx$ .

$I = \int {{{\sin }^2}\left( {2x + 1} \right) \cdot \sin \left( {2x + 1} \right)} dx$

As it is not integrable directly, therefore we will use trigonometry formula which is

${\sin ^2}\theta = 1 - {\cos ^2}\theta$

Therefore, above equation becomes,

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

Integrating w.r.t. $x$ .

$\int y dx = \int {\left( {1 - {{\cos }^2}\left( {2x + 1} \right)} \right).\sin \left( {2x + 1} \right)} dx$……$\left( 1 \right)$ .

Now, using substitution method,

Let $\cos \left( {2x + 1} \right) = t$

Differentiating w.r.t. $x$ .

$\cos \left( {2x + 1} \right) = t \hfill \\ - 2\sin \left( {2x + 1} \right) = \dfrac{{dt}}{{dx}} \hfill \\ \sin \left( {2x + 1} \right)dx = - \dfrac{{dt}}{2} \hfill \\$

Substituting the value of $\sin \left( {2x + 1} \right)dx$ in equation $\left( 1 \right)$ .

Therefore,

$I = - \dfrac{1}{2}\int {\left( {1 - {t^2}} \right)} dt \hfill \\ I = - \dfrac{1}{2}\left\{ {t - \dfrac{{{t^3}}}{3}} \right\} + C \hfill \\$

Substituting the value of $t$,

So,

$I = - \dfrac{1}{2}\left\{ {\cos \left( {2x + 1} \right) - \dfrac{{{{\cos }^3}\left( {2x + 1} \right)}}{3}} \right\} + C$

Where $C$ is an arbitrary constant.

5. Integrate ${\sin ^3}x{\cos ^3}x$ .

Ans: let $I = \int {{{\sin }^3}x{{\cos }^3}x} dx$ .

$I = \int {{{\cos }^3}x{{\sin }^2}x\sin xdx}$

As it is not integrable directly, therefore we will use trigonometry formula which is

${\sin ^2}x = 1 - {\cos ^2}x$

Therefore,

$I = \int {{{\cos }^3}x\left( {1 - {{\cos }^2}x} \right)\sin xdx}$ ……$\left( 1 \right)$

Let $\cos x = t$ .

Differentiating w.r.t. $x$ .

$- \sin x = \dfrac{{dt}}{{dx}} \hfill \\ \sin xdx = - dt \hfill \\$

Substituting the value of $\sin xdx$in $\left( 1 \right)$ .

$I = - \int {{t^3}\left( {1 - {t^2}} \right)dt}\\ = - \int {\left( {{t^3} - {t^5}} \right)dt}\\ = - \left\{ {\dfrac{{{t^4}}}{4} - \dfrac{{{t^6}}}{6} + C} \right\} \\$

Now, substituting the value of $t$,

$I = \dfrac{{{{\cos }^6}x}}{6} - \dfrac{{{{\cos }^4}x}}{4} + C$

Where $C$ is an arbitrary constant.

6. integrate $\sin x\sin 2x\sin 3x$.

Ans:  To solve this problem we will be using the following formula

$\begin{gathered} \sin A\sin B = \frac{1}{2}\left\{ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right\} \\ \sin A\cos B = \frac{1}{2}\left\{ {\sin \left( {A + B} \right) + \sin \left( {A - B} \right)} \right\} \\ \sin 2A = 2\sin A\cos A \\ \end{gathered}$

Therefore, on simplifying the given equation,

$\begin{gathered} \int {\sin x\sin 2x\sin 3xdx} = \int {\left[ {\sin x.\frac{1}{2}\left\{ {\cos \left( {2x - 3x} \right) - \cos \left( {2x + 3x} \right)} \right\}} \right]} dx \\ = \frac{1}{2}\int {\left( {\sin x\cos \left( { - x} \right) - \sin x\cos 5x} \right)} dx \\ = \frac{1}{2}\int {\left( {\sin x\cos x - \sin x\cos 5x} \right)} dx \\ = \frac{1}{2}\int {\frac{{\sin 2x}}{2}dx - \frac{1}{2}\int {\sin x\cos 5xdx} }\\ = \frac{1}{4}\left[ {\frac{{ - \cos 2x}}{2}} \right] - \frac{1}{2}\int {\frac{1}{2}\left\{ {\sin \left( {x + 5x} \right) + \sin \left( {x - 5x} \right)} \right\}dx}\\ = \frac{{ - \cos 2x}}{8} - \frac{1}{4}\int {\left( {\sin 6x + \sin \left( { - 4x} \right)} \right)dx}\\ = \frac{{ - \cos 2x}}{8} - \frac{1}{4}\int {\left( {\sin 6x - \sin 4x} \right)dx}\\ = \frac{{ - \cos 2x}}{8} - \frac{1}{4}\left[ {\frac{{ - \cos 6x}}{6} + \frac{{\cos 4x}}{4}} \right] + C \\ = \frac{1}{8}\left[ {\frac{{\cos 6x}}{3} - \frac{{\cos 4x}}{4} - \cos 2x} \right] + C \\ \end{gathered}$

Where $C$ is an arbitrary constant.

7. Integrate $\sin 4x\sin 8x$.

Ans:  to solve this problem we will be using the formula,

$\sin A\sin B = \dfrac{1}{2}\left\{ {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right\}$

Therefore, simplifying the given equation,

$\int {\sin 4x\sin 8xdx} = \int {\left[ {\dfrac{1}{2}\left\{ {\cos \left( {4x - 8x} \right) - \cos \left( {4x + 8x} \right)} \right\}} \right]} dx \\ = \dfrac{1}{2}\int {\left( {\cos \left( { - 4x} \right) - \cos 12x} \right)} dx \\ = \dfrac{1}{2}\int {\left( {\cos 4x - \cos 12x} \right)} dx \\ = \dfrac{1}{2}\left[ {\dfrac{{\sin 4x}}{4} - \dfrac{{\sin 12x}}{{12}}} \right] + C \\$

Where $C$ is an arbitrary constant.

8. Integrate $\dfrac{{1 - \cos x}}{{1 + \cos x}}$ .

Ans:  Using the formula,

$1 - \cos x = 2{\sin ^2}\dfrac{x}{2}{\text{ and }}1 + \cos x = 2{\cos ^2}\dfrac{x}{2}$

Therefore, simplifying the given equation,

$\dfrac{{1 - \cos x}}{{1 + \cos x}} = \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}} \\ = {\tan ^2}\dfrac{x}{2} \\ = {\sec ^2}\dfrac{x}{2} - 1 \\$

Therefore,

$\int {\dfrac{{1 - \cos x}}{{1 + \cos x}}dx} = \int {\left( {{{\sec }^2}\dfrac{x}{2} - 1} \right)} dx \\ = 2\tan \dfrac{x}{2} - x + C \\$

Where $C$ is an arbitrary constant.

9. Integrate $\dfrac{{\cos x}}{{1 + \cos x}}$.

Ans:  Using the formula,

$\cos x = {\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}{\text{ and }}1 + \cos x = 2{\cos ^2}\dfrac{x}{2}$

Therefore, simplifying the given equation,

$\dfrac{{\cos x}}{{1 + \cos x}} = \dfrac{{{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}} \\ = \dfrac{1}{2} - \dfrac{1}{2}{\tan ^2}\dfrac{x}{2} \\ = \dfrac{1}{2} - \dfrac{1}{2}\left( {{{\sec }^2}\dfrac{x}{2} - 1} \right) \\ = 1 - \dfrac{1}{2}{\sec ^2}\dfrac{x}{2} \\$

Therefore,

$\int {\dfrac{{\cos x}}{{1 + \cos x}}} = \int {\left( {1 - \dfrac{1}{2}{{\sec }^2}\dfrac{x}{2}} \right)} dx \\ = x - 2 \times \dfrac{1}{2}\tan \dfrac{x}{2} + C \\ = x - \tan \dfrac{x}{2} + C \\$

Where $C$ is an arbitrary constant.

10. Integrate ${\sin ^4}x$.

Ans:  Using the formula,

${\sin ^2}x = \frac{{1 - \cos 2x}}{2}{\text{ and co}}{{\text{s}}^2}2x = \frac{{1 + \cos 4x}}{2}$

Therefore, simplifying the given equation,

$\begin{gathered} {\sin ^4}x = {\sin ^2}x{\sin ^2}x \\ = \left( {\frac{{1 - \cos 2x}}{2}} \right)\left( {\frac{{1 - \cos 2x}}{2}} \right) \\ = \frac{1}{4}{\left( {1 - \cos 2x} \right)^2} \\ = \frac{1}{4}\left[ {1 + {{\cos }^2}2x - 2\cos 2x} \right] \\ = \frac{1}{4}\left[ {1 + \left( {\frac{{1 + \cos 4x}}{2}} \right) - 2\cos 2x} \right] \\ = \frac{1}{4}\left[ {1 + \frac{1}{2} + \frac{{\cos 4x}}{2} - 2\cos 2x} \right] \\ = \frac{1}{4}\left[ {\frac{3}{2} + \frac{{\cos 4x}}{2} - 2\cos 2x} \right] \\ \end{gathered}$

Therefore,

$\begin{gathered} \int {{{\sin }^4}x} = \frac{1}{4}\int {\left[ {\frac{3}{2} + \frac{{\cos 4x}}{2} - 2\cos 2x} \right]} dx \\ = \frac{1}{4}\left[ {\frac{3}{2}x + \frac{{\sin 4x}}{8} - \frac{{2\sin 2x}}{2}} \right] + C \\ = \frac{1}{4}\left[ {\frac{3}{2}x + \frac{{\sin 4x}}{8} - \sin 2x} \right] + C \\ = \frac{3}{8}x + \frac{{\sin 4x}}{{32}} - \frac{{\sin 2x}}{4} + C \\ \end{gathered}$

Where $C$ is an arbitrary constant.

11. Integrate ${\cos ^4}4x$.

Ans:  Using the formula,

${\cos ^2}2x = \dfrac{{1 + \cos 4x}}{2}$

Therefore, simplifying the given equation,

${\cos ^4}2x = {\left( {{{\cos }^2}2x} \right)^2} \\ = {\left( {\dfrac{{1 + \cos 4x}}{2}} \right)^2} \\ = \dfrac{1}{4}\left[ {1 + {{\cos }^2}4x + 2\cos 4x} \right] \\ = \dfrac{1}{4}\left[ {1 + \dfrac{{1 + \cos 8x}}{2} + 2\cos 4x} \right] \\ = \dfrac{1}{4}\left[ {1 + \dfrac{1}{2} + \dfrac{{\cos 8x}}{2} + 2\cos 4x} \right] \\ = \dfrac{1}{4}\left[ {\dfrac{3}{2} + \dfrac{{\cos 8x}}{2} + 2\cos 4x} \right] \\$

Therefore,

$\int {{{\cos }^4}2xdx} = \dfrac{1}{4}\int {\left[ {\dfrac{3}{2} + \dfrac{{\cos 8x}}{2} + 2\cos 4x} \right]} dx \\ = \dfrac{1}{4}\left[ {\dfrac{3}{2}x + \dfrac{{\sin 8x}}{{16}} + \dfrac{{2\sin 4x}}{4}} \right] + C \\ = \dfrac{3}{8}x + \dfrac{{\sin 8x}}{{64}} + \dfrac{{\sin 4x}}{8} + C \\$

Where $C$ is an arbitrary constant.

12. Integrate $\dfrac{{{{\sin }^2}x}}{{1 + \cos x}}$.

Ans:  Using the formula,

$\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \hfill \\ {\text{1 + }}\cos x = 2{\cos ^2}\dfrac{x}{2} \hfill \\ 1 - \cos x = 2{\sin ^2}\dfrac{x}{2} \hfill \\$

Therefore, simplifying the given equation,

$\dfrac{{{{\sin }^2}x}}{{1 + \cos x}} = \dfrac{{{{\left( {2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right)}^2}}}{{2{{\cos }^2}\dfrac{x}{2}}} \\ = 2{\sin ^2}\dfrac{x}{2} \\ = 1 - \cos x \\$

Therefore,

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

Where $C$ is an arbitrary constant.

13. Integrate $\dfrac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}$.

Ans:  Using the formula,

$\cos C - \cos D = - 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right) \hfill \\ \sin A = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2} \hfill \\$

Therefore, simplifying the given equation,

$\dfrac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }} = \dfrac{{ - 2\sin \left( {\dfrac{{2x + 2\alpha }}{2}} \right)\sin \left( {\dfrac{{2x - 2\alpha }}{2}} \right)}}{{ - 2\sin \left( {\dfrac{{x + \alpha }}{2}} \right)\sin \left( {\dfrac{{x - \alpha }}{2}} \right)}} \\ = \dfrac{{\sin \left( {x + \alpha } \right)\sin \left( {x - \alpha } \right)}}{{\sin \left( {\dfrac{{x + \alpha }}{2}} \right)\sin \left( {\dfrac{{x - \alpha }}{2}} \right)}} \\ = \dfrac{{\left[ {2\sin \left( {\dfrac{{x + \alpha }}{2}} \right)\cos \left( {\dfrac{{x + \alpha }}{2}} \right)} \right]\left[ {2\sin \left( {\dfrac{{x - \alpha }}{2}} \right)\cos \left( {\dfrac{{x - \alpha }}{2}} \right)} \right]}}{{\sin \left( {\dfrac{{x + \alpha }}{2}} \right)\sin \left( {\dfrac{{x - \alpha }}{2}} \right)}} \\ = 4\cos \left( {\dfrac{{x + \alpha }}{2}} \right)\cos \left( {\dfrac{{x - \alpha }}{2}} \right) \\ = 2\left[ {\cos \left( {\dfrac{{x + \alpha }}{2} + \dfrac{{x - \alpha }}{2}} \right) + \cos \left( {\dfrac{{x + \alpha }}{2} - \dfrac{{x - \alpha }}{2}} \right)} \right] \\ = 2\left( {\cos x + \cos \alpha } \right) \\$

Therefore,

$\int {\left( {\dfrac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}} \right)} dx = = 2\int {\left( {\cos x + \cos \alpha } \right)} dx \\ = 2\left[ {\sin x + x\cos \alpha } \right] + C \\$

Where $C$ is an arbitrary constant.

14. Integrate $\dfrac{{\cos x - \sin x}}{{1 + \sin 2x}}$.

Ans:  Using the formula,

${\sin ^2}x + {\cos ^2}x = 1 \\ \sin 2x = 2\sin x\cos x \\$

Therefore, simplifying the given equation,

$\dfrac{{\cos x - \sin x}}{{1 + \sin 2x}} = \dfrac{{\cos x - \sin x}}{{{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x}} \\ = \dfrac{{\cos x - \sin x}}{{{{\left( {\sin x + \cos x} \right)}^2}}} \\$

Therefore,

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

Let $\sin x + \cos x = t$

Differentiating w.r.t. $x$,

$\cos x - \sin x = \dfrac{{dt}}{{dx}} \\ \left( {\cos x - \sin x} \right)dx = dt \\$

Substituting the value of $\left( {\cos x - \sin x} \right)dx = dt$ in equation $\left( 1 \right)$ .

Therefore,

$I = \int {\dfrac{1}{{{t^2}}}} dt \\ = - \dfrac{1}{t} + C \\$

Substituting the value of $t$,

Therefore,

$I = - \dfrac{1}{{\sin x + \cos x}} + C$

Where $C$ is an arbitrary constant.

15. Integrate ${\tan ^3}2x\sec 2x$.

Ans:  Using the formula,

${\sec ^2}x - 1 = {\tan ^2}x$

Therefore,

${\tan ^3}2x\sec 2x = \left( {{{\sec }^2}2x - 1} \right)\tan 2x\sec 2x \\ = {\sec ^2}2x\tan 2x\sec 2x - \tan 2x\sec 2x \\$

Therefore,

$\int {{{\tan }^3}2x\sec 2xdx} = \int {{{\sec }^2}2x\tan 2x\sec 2x} dx - \int {\tan 2x\sec 2x} dx \\ = \int {{{\sec }^2}2x\tan 2x\sec 2x} dx - \dfrac{{\sec 2x}}{2} + C{\text{ }}......\left( 1 \right) \\$

Let $\sec 2x = t$

Differentiating w.r.t. $x$,

$\sec 2x = t \\ 2\sec 2x\tan 2x = \dfrac{{dt}}{{dx}} \\ \sec 2x\tan 2xdx = \dfrac{{dt}}{2} \\$

Substituting the value of $\sec 2x\tan 2xdx = \dfrac{{dt}}{2}$ in equation $\left( 1 \right)$ .

$I = \int {{{\sec }^2}2x\tan 2x\sec 2x} dx - \dfrac{{\sec 2x}}{2} + C \\ = \dfrac{1}{2}\int {{t^2}dt - } \dfrac{{\sec 2x}}{2} + C \\ = \dfrac{1}{2}\left( {\dfrac{{{t^3}}}{3}} \right) - \dfrac{{\sec 2x}}{2} + C \\$

Substituting the value of $t$,

$I = \dfrac{{{{\sec }^3}2x}}{6} - \dfrac{{\sec 2x}}{2} + C$

Where $C$ is an arbitrary constant.

16. Integrate ${\tan ^4}x$.

Ans:  Using the formula,

${\sec ^2}x - 1 = {\tan ^2}x$

Therefore, simplifying the given equation,

${\tan ^4}x = {\tan ^2}x{\tan ^2}x \\ = \left( {{{\sec }^2}x - 1} \right){\tan ^2}x \\ = {\sec ^2}x{\tan ^2}x - {\tan ^2}x \\ = {\sec ^2}x{\tan ^2}x - \left( {{{\sec }^2}x - 1} \right) \\ = {\sec ^2}x{\tan ^2}x - {\sec ^2}x + 1 \\$

Therefore,

$\int {{{\tan }^4}x} dx = \int {\left( {{{\sec }^2}x{{\tan }^2}x - {{\sec }^2}x + 1} \right)dx}\\ = \int {{{\sec }^2}x{{\tan }^2}xdx} - \int {{{\sec }^2}xdx + \int {1.dx} }\\ = \int {{{\sec }^2}x{{\tan }^2}xdx} - \tan x + x + C_1{\text{ }}......\left( 1 \right) \\$

Consider, $I = \int {{{\sec }^2}x{{\tan }^2}xdx}$

Let $\tan x = t$

Differentiating w.r.t. $x$,

$\tan x = \dfrac{{dt}}{{dx}} \\ {\sec ^2}xdx = dt \\$

Substituting the value of ${\sec ^2}xdx = dt$ in above equation,

Therefore,

$I = \int {{t^2}dt}\\ = \dfrac{{{t^3}}}{3} + C_2 \\$

Substituting the value of $t$,

Therefore,

$\int {{{\sec }^2}x{{\tan }^2}xdx} = \dfrac{{{{\tan }^3}x}}{3} + C_2$

Substituting this value in equation $\left( 1 \right)$.

Therefore,

$\int {{{\tan }^4}x} dx = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C_1+C_2$

$\int {{{\tan }^4}x} dx = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C$

Where $C=C_1+C_2$ , is an arbitrary constant.

17. Integrate $\dfrac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}$.

Ans:  simplifying the given equation,

$\dfrac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}} = \dfrac{{{{\sin }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}} + \dfrac{{{{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}} \\ = \dfrac{{\sin x}}{{{{\cos }^2}x}} + \dfrac{{\cos x}}{{{{\sin }^2}x}} \\ = \tan x\sec x + \cot x\cos ecx \\$

Therefore,

$\int {\left( {\dfrac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}} \right)} dx = \int {\left( {\tan x\sec x + \cot x\cos ecx} \right)} dx \\ = \sec x - \cos ecx + C \\$

Where $C$ is an arbitrary constant.

18. Integrate $\dfrac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}$.

Ans:  using the formula,

$\cos 2x = 1 - 2{\sin ^2}x$

Therefore, simplifying the given equation,

$\dfrac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}} = \dfrac{{\cos 2x + \left( {1 - \cos 2x} \right)}}{{{{\cos }^2}x}} \\ = \dfrac{1}{{{{\cos }^2}x}} \\ = {\sec ^2}x \\$

Therefore,

$\int {\left( {\dfrac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}} \right)} dx = \int {{{\sec }^2}x} dx \\ = \tan x + C \\$

Where $C$ is an arbitrary constant.

19. Integrate $\dfrac{1}{{\sin x{{\cos }^3}x}}$.

Ans:  Using the formula,

${\cos ^2}x + {\sin ^2}x = 1$

Therefore, simplifying the given equation,

$\dfrac{1}{{\sin x{{\cos }^3}x}} = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\sin x{{\cos }^3}x}} \\ = \dfrac{{{{\cos }^2}x}}{{\sin x{{\cos }^3}x}} + \dfrac{{{{\sin }^2}x}}{{\sin x{{\cos }^3}x}} \\ = \dfrac{1}{{\sin x\cos x}} + \dfrac{{\sin x}}{{{{\cos }^3}x}} \\ = \dfrac{{\dfrac{1}{{{{\cos }^2}x}}}}{{\dfrac{{\sin x\cos x}}{{{{\cos }^2}x}}}} + \tan x{\sec ^2}x \\ = \dfrac{{{{\sec }^2}x}}{{\tan x}} + \tan x{\sec ^2}x \\$

Therefore,

$\int {\dfrac{1}{{\sin x{{\cos }^3}x}}} dx = \int {\dfrac{{{{\sec }^2}x}}{{\tan x}}} dx + \int {\tan x{{\sec }^2}x} dx$

Let $\tan x = t$,

Differentiating w.r.t. $x$,

$\tan x = \dfrac{{dt}}{{dx}} \\ {\sec ^2}xdx = dt \\$

Substituting the value of ${\sec ^2}xdx = dt$ in above equation,

Therefore,

$\int {\dfrac{1}{{\sin x{{\cos }^3}x}}} dx = \int {\dfrac{1}{t}} dt + \int t dt \\ = \log \left| t \right| + \dfrac{{{t^2}}}{2} + C \\$

Substituting the value of $t$,

$\int {\dfrac{1}{{\sin x{{\cos }^3}x}}} dx = \log \left| {\tan x} \right| + \dfrac{{{{\tan }^2}x}}{2} + C$

Where $C$ is arbitrary constant.

20. Integrate $\dfrac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}$

Ans:  On simplifying the equation,

$\begin{gathered} \frac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}} = \frac{{\cos 2x}}{{{{\cos }^2}x + {{\sin }^2}x + 2\sin 2x}} \\ = \frac{{\cos 2x}}{{1 + \sin 2x}} \\ \end{gathered}$

Therefore,

$\int {\frac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx} = \int {\frac{{\cos 2x}}{{1 + \sin 2x}}} dx$

Let $1 + \sin 2x = t$

Differentiating w.r.t. $x$,

$\begin{gathered} 2\cos 2x = \frac{{dt}}{{dx}} \hfill \\ \cos 2xdx = \frac{{dt}}{2} \hfill \\ \end{gathered}$

Substituting the value of $\cos 2xdx = \frac{{dt}}{2}$ in above equation,

Therefore,

$\begin{gathered} \int {\frac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx} = \frac{1}{2}\int {\frac{1}{t}dt}\\ = \frac{1}{2}\log \left| t \right| + C \\ \end{gathered}$

Substituting the value of $t$,

$\begin{gathered} \int {\frac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx} = \frac{1}{2}\log \left| {1 + 2\sin 2x} \right| + C \\ = \frac{1}{2}\log \left| {{{\cos }^2}x + {{\sin }^2}x + 2\sin 2x} \right| + C \\ = \frac{1}{2}\log \left| {{{\left( {\cos x + \sin x} \right)}^2}} \right| + C \\ = \log \left| {\cos x + \sin x} \right| + C \\ \end{gathered}$

Where $C$ is an arbitrary constant.

21. Integrate ${\sin ^{ - 1}}\left( {\cos x} \right)$.

Ans:  Using the formula,

${\cos ^2}x + {\sin ^2}x = 1$

Let $\cos x = t$…… $\left( 1 \right)$

Therefore,

$\sin x = \sqrt {1 - {t^2}}$

Differentiating equation $\left( 1 \right)$ w.r.t. $x$,

$- \sin x = \dfrac{{dt}}{{dx}} \hfill \\ dx = - \dfrac{{dt}}{{\sin x}} \hfill \\ dx = - \dfrac{{dt}}{{\sqrt {1 - {t^2}} }} \hfill \\$

Therefore, the given question becomes,

$\int {{{\sin }^{ - 1}}\left( {\cos x} \right)dx = } \int {{{\sin }^{ - 1}}t\left( {\dfrac{{ - dt}}{{\sqrt {1 - {t^2}} }}} \right)}$

$\int {{{\sin }^{ - 1}}\left( {\cos x} \right)dx = } - \int {\dfrac{{{{\sin }^{ - 1}}t}}{{\sqrt {1 - {t^2}} }}dt}$……$\left( 1 \right)$

Let ${\sin ^{ - 1}}t = u$

Differentiating w.r.t. $t$,

$\dfrac{1}{{\sqrt {1 - {t^2}} }}dt = du$

Therefore, equation becomes,

$\int {{{\sin }^{ - 1}}\left( {\cos x} \right)dx} = - \int {udu}\\ = - \dfrac{{{u^2}}}{2} + C \\ = - \dfrac{{{{\left( {{{\sin }^{ - 1}}t} \right)}^2}}}{2} + C \\ = - \dfrac{{{{\left( {{{\sin }^{ - 1}}\cos x} \right)}^2}}}{2} + C......\left( 2 \right) \\$

As we know that,

${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$

Therefore,

${\sin ^{ - 1}}\left( {\cos x} \right) = \dfrac{\pi }{2} - {\cos ^{ - 1}}\left( {\cos x} \right) \\ = \dfrac{\pi }{2} - x \\$

Substituting this value is equation $\left( 2 \right)$

Therefore,

$\int {{{\sin }^{ - 1}}\left( {\cos x} \right)dx} = - \dfrac{{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}{2} + C \\ = - \dfrac{1}{2}\left( {\dfrac{{{\pi ^2}}}{4} + {x^2} - \pi x} \right) + C \\ = - \dfrac{{{\pi ^2}}}{8} - \dfrac{{{x^2}}}{2} + \dfrac{{\pi x}}{2} + C \\ = \dfrac{{\pi x}}{2} - \dfrac{{{x^2}}}{2} + \left( {C - \dfrac{{{\pi ^2}}}{8}} \right) \\ = \dfrac{{\pi x}}{2} - \dfrac{{{x^2}}}{2} + C_1 \\$

Where $C_1$ is an arbitrary constant.

22. Integrate $\dfrac{1}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}$.

Ans:  Using the formula,

$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$

Therefore, simplifying the given question,

$\dfrac{1}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}} = \dfrac{1}{{\sin \left( {a - b} \right)}}\left[ {\dfrac{{\sin \left( {a - b} \right)}}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}} \right] \\ = \dfrac{1}{{\sin \left( {a - b} \right)}}\left[ {\dfrac{{\sin \left( {a - b - x + x} \right)}}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}} \right] \\ = \dfrac{1}{{\sin \left( {a - b} \right)}}\left[ {\dfrac{{\sin \left[ {\left( {x - b} \right) - \left( {x - a} \right)} \right]}}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}} \right] \\ = \dfrac{1}{{\sin \left( {a - b} \right)}}\left[ {\dfrac{{\sin \left( {x - b} \right)\cos \left( {x - a} \right) - \cos \left( {x - b} \right)\sin \left( {x - a} \right)}}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}} \right] \\ = \dfrac{1}{{\sin \left( {a - b} \right)}}\left[ {\tan \left( {x - b} \right) - \tan \left( {x - a} \right)} \right] \\$

Therefore,

$\int {\dfrac{1}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}} dx = \dfrac{1}{{\sin \left( {a - b} \right)}}\int {\left[ {\tan \left( {x - b} \right) - \tan \left( {x - a} \right)} \right]} dx \\ = \dfrac{1}{{\sin \left( {a - b} \right)}}\left[ { - \log \left| {\cos \left( {x - b} \right)} \right| + \log \left| {\cos \left( {x - a} \right)} \right|} \right] + C \\ = \dfrac{1}{{\sin \left( {a - b} \right)}}\left[ {\log \left| {\dfrac{{\cos \left( {x - a} \right)}}{{\cos \left( {x - b} \right)}}} \right|} \right] + C \\$

Where $C$ is an arbitrary constant.

23. $\int {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}}$ is equal to

$\left( A \right)\tan x + \cot x + C{\text{ }}\left( B \right)\tan x + \cos ecx + C \hfill \\ \left( C \right) - \tan x + \cot x + C{\text{ }}\left( C \right)\tan + \sec x + C \hfill \\$

Ans:

$\int {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx = \int {\left( {\dfrac{{{{\sin }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} - \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} \right)} dx \\ = \int {\left( {{{\sec }^2}x - \cos e{c^2}x} \right)dx}\\ = \tan x + \cot x + C \\$

Therefore, the correct option is $\left( A \right)$.

24. $\int {\dfrac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {{e^x}x} \right)}}dx}$equals

$\left( A \right) - \cot \left( {x{e^x}} \right) + C{\text{ }}\left( B \right)\tan \left( {x{e^x}} \right)C \hfill \\ \left( C \right)\tan \left( {{e^x}} \right) + C{\text{ }}\left( C \right)\cot \left( {{e^x}} \right) + C \hfill \\$

Ans: $\int {\dfrac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {{e^x}x} \right)}}dx}$

Let ${e^x}x = t$

Differentiating w.r.t. $x$,

${e^x}.x + {e^x}.1 = \dfrac{{dt}}{{dx}} \\ {e^x}\left( {1 + x} \right)dx = dt \\$

Therefore, the given question becomes,

$\int {\dfrac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {{e^x}x} \right)}}dx} = \int {\dfrac{{dt}}{{{{\cos }^2}t}}}\\ = \int {{{\sec }^2}tdt}\\ = \tan t + C \\ = \tan \left( {{e^x}x} \right) + C \\$

Therefore, the correct option is $\left( B \right)$.

## What is There in Exercise 7.6 Class 12 NCERT Solutions?

Math is a complicated subject, and so is integration. Unless you understand all the steps correctly, you won't solve the problems. If you are a complete beginner in integration or face considerable difficulties in Ex 7.6 Class 12 Maths NCERT, then we have the ultimate solution for your understanding. The best guide for you is Exercise 7.6 Class 12 NCERT Solutions if you want to learn integrals. The content of the NCERT book is as per the latest CBSE syllabus and of high quality. However, the language is straightforward and concise so that the reader grasps it quickly. The matter available in the book is entirely reliable.

The easy language makes the book more viable and user-friendly. 7.1 is about the introduction. Again, 7.2 is about integration through the reverse procedure of differentiation. While 7.3 discusses the different methods of integration. 7.4 has integrals of specific functions, and 7.5 has integration with partial functions. NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.6 integration by parts. 7.7 and 7.8 have definite integral and fundamental calculus theorem respectively. 7.9 talks about evaluating definite integrals with substitution method and 7.10 have some essential properties of definite integrals.

### What Is Integration by Parts of Ex 7.6?

Integration by parts usually teaches a student to solve the product's integration of any two or more functions. It is also known as the partial integration method. Here, we can write d(fa fb)/dx = fa d(fb)/dx + fb d(fa)/dx, where fa and fb are two different functions for a singular variable x. Here, you will also learn how to apply the integral on both sides. Finally, the product of both functions can be integrated by (first function) * (second function integral) – integral of ( first function of differential coefficient) * (second function on integration).

In all Class 12 Maths NCERT Solutions Chapter 7 Exercise 7.6 deals with displacement and volume by putting numbers into functions. Integrals are one of the main applications of calculus, which was hand-in-hand with differentiation and inverse proportion. Every problem clears your concepts in detail. Hence, one should continue to practice every day for good results. Maths require continuous hard work and practice.

### How Class 12 Maths Chapter 7 Exercise 7.6 Help Students?

Students find Exercise 7.6 Class 12 NCERT solutions helpful because the content is written by expert teachers and professionals who have extensive experience in the education field. Teachers from the entire country and different parts of the world give their input to create unique content on Class 12 Maths ch 7 Ex 7.6. There are times when textbooks are not enough to help you clear your concepts. Here you need external help. While Class 12 Maths 7.6 Solutions allow one to earn their ideas for better understanding. Instead of attending tuitions at coaching centres, wasting money and time, you can get all the study materials at your fingertips. Here you can get the complete authentic Class 12 Maths Chapter 7 Exercise 7.6 solutions. It helps boost the student's confidence before exams and understand how to manage their time. You get to know which questions you have to practice more.

### Solved Examples

Question1: Why is Class 12 Calculus Complicated?

Answer: Class 12 Calculus is difficult for beginners. However, once you understand the chapter, it becomes one of the most exciting chapters to solve.

Question2: What does NCERT Solutions Maths Class 12 Chapter 7 Cover?

Answer: Maths Class 12 Chapter 7 is all about integrals, partial fraction, integration, integration by parts and other forms.

Q1. What are the Advantages of Ex 7.6 Class 12 Maths NCERT Solutions?

Answer: Not only study material, but you also have extra exercises to practice. The first step to do is read the entire Ex 7.6 Class 12 Maths Solutions and learn the concepts. Then you can easily copy down some of the solved examples from the material. All the sums are done following a good number of steps. If you don't follow the same steps for exams, then you will not score good marks. In Ex 7.6 Class 12 Maths Solutions, there is step marking. Hence you need to do all the correct steps and get the proper answer to achieve full marks. Using shortcut methods to arrive at the final answer somehow won't fetch you marks. There are 24 solutions with questions in Maths NCERT Solutions Class 12 Chapter 7 Exercise 7.6. Other activities also have solutions.

Q2. How to Differentiate Between Indefinite and Definite Integrals from Maths NCERT Class 12 Ex 7.6 Solutions?

Answer: Definite integrals are those which have both lower and upper limits given for an integral. In reality, you are calculating the area under the curve of x = c and x = d for function f(x) with a lower limit as c and an upper limit as b. For indefinite integrals, no upper or lower limits are provided. Here x can have a range of solutions because you calculate a general solution for a family of the same functions like f(x). In indefinite integrals, we need to add a fixed constant c, and the notation is written along with the solution we get. So this is the fundamental difference between definite and indefinite integrals of Exercise 7.6 Class 12 Maths NCERT solutions. SHARE TWEET SHARE SUBSCRIBE