# NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3

Go back to  'Integrals'

## Chapter 7 Ex.7.3 Question 1

Find the integral of $${\sin ^2}\left( {2x + 5} \right)$$

### Solution

\begin{align} &{\sin ^2}\left( {2x + 5} \right)= \frac{{1 - \cos 2\left( {2x + 5} \right)}}{2} = \frac{{1 - \cos \left( {4x + 10} \right)}}{2}\\ & \Rightarrow \;\int {{{\sin }^2}} \left( {2x + 5} \right)dx = \int {\frac{{1 - \cos \left( {4x + 10} \right)}}{2}dx} \\ & = \frac{1}{2}\int {1dx - \frac{1}{2}\int {\cos \left( {4x + 10} \right)dx} } \\ & = \frac{1}{2}x - \frac{1}{2}\left( {\frac{{\sin \left( {4x + 10} \right)}}{4}} \right) + C\\ & = \frac{1}{2}x - \frac{1}{8}\sin \left( {4x + 10} \right) + C\end{align}

## Chapter 7 Ex.7.3 Question 2

Find the integral of $$\sin 3x\cos 4x$$

### Solution

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

\begin{align}&\therefore \int {\sin 3x\cos 4xdx = \frac{1}{2}\int {\left\{ {\sin \left( {3x + 4x} \right) + \sin \left( {3x - 4x} \right)} \right\}dx} } \\ &= \frac{1}{2}\int {\left\{ {\sin 7x + \sin \left( { - x} \right)} \right\}} dx\\ &= \frac{1}{2}\int {\left\{ {\sin 7x - \sin x} \right\}dx} \\& = \frac{1}{2}\int {\sin 7xdx - \frac{1}{2}} \int {\sin xdx} \\& = \frac{1}{2}\left( {\frac{{ - \cos 7x}}{7}} \right) - \frac{1}{2}\left( { - \cos x} \right) + C\\& = \frac{{ - \cos 7x}}{{14}} + \frac{{\cos x}}{2} + C\end{align}

## Chapter 7 Ex.7.3 Question 3

Find the integral of $$\cos 2x\cos 4x\cos 6x$$

### Solution

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

\begin{align}&\therefore \int {\cos 2x\left( {\cos 4x\cos 6x} \right)dx = \int {\cos 2x\left[ {\frac{1}{2}\left\{ {\cos \left( {4x + 6x} \right) + \cos \left( {4x - 6x} \right)} \right\}} \right]dx} } \\ &= \frac{1}{2}\int {\left\{ {\cos 2x\cos 10x + \cos 2x\cos \left( { - 2x} \right)} \right\}} dx\\ &= \frac{1}{2}\int {\left\{ {\cos 2x\cos 10x + {{\cos }^2}2x} \right\}dx} \\ &= \frac{1}{2}\int {\left[ {\left\{ {\frac{1}{2}\cos \left( {2x + 10x} \right) + \frac{1}{2}\cos \left( {2x - 10x} \right)} \right\} + \left( {\frac{{1 + \cos 4x}}{2}} \right)} \right]dx} \\& = \frac{1}{4}\int {\left( {\cos 12x + \cos 8x + 1 + \cos 4x} \right)dx} \\& = \frac{1}{4}\left[ {\frac{{\sin 12x}}{{12}} + \frac{{\sin 8x}}{8} + x + \frac{{\sin 4x}}{4} + C} \right]\end{align}

## Chapter 7 Ex.7.3 Question 4

Find the integral of $${\sin ^3}\left( {2x + 1} \right)$$

### Solution

Put, I$$\; = \int {{{\sin }^3}\left( {2x + 1} \right)}$$

\begin{align} \qquad \Rightarrow \;\int {{{\sin }^3}\left( {2x + 1} \right)dx = \int {{{\sin }^2}\left( {2x + 1} \right)\sin \left( {2x + 1} \right)dx} } \\ \qquad\qquad\qquad = \int {\left( {1 - {{\cos }^2}\left( {2x + 1} \right)} \right)\sin \left( {2x + 1} \right)dx}\end{align}

\begin{align}&{\rm{Let }}\cos \left( {2x + 1} \right) = t\\[5pt]& \Rightarrow \;- 2\sin \left( {2x + 1} \right)dx = dt\\ &\Rightarrow \;\sin \left( {2x + 1} \right)dx =\frac{{ - dt}}{2}\\ \Rightarrow \;{\rm{I}} &= \frac{{ - 1}}{2}\int {\left( {1 - {t^2}} \right)dt} \\ &= \frac{{ - 1}}{2}\left\{ {t - \frac{{{t^3}}}{3}} \right\}\\& = \frac{{ - 1}}{2}\left\{ {\cos \left( {2x + 1} \right) - \frac{{{{\cos }^3}\left( {2x + 1} \right)}}{3}} \right\}\\ &= \frac{{ - \cos \left( {2x + 1} \right)}}{2} + \frac{{{{\cos }^3}\left( {2x + 1} \right)}}{6} + C\end{align}

## Chapter 7 Ex.7.3 Question 5

Find the integral of $${\sin ^3}x\;{\cos ^3}x$$

### Solution

Let I$$= \int {{{\sin }^3}x\;{{\cos }^3}x\;dx}$$

\begin{align}& = \int {{{\cos }^3}x\;{{\sin }^2}x\;\sin x\;dx} \\ &= \int {{{\cos }^3}x\left( {1 - {{\cos }^2}x} \right)\sin x\;dx}\end{align}

Let $$\cos x = t$$

\begin{align}& \Rightarrow \;- \sin x\;dx = dt\\& \Rightarrow \;{\rm{I}} = - \int {{t^3}\left( {1 - {t^2}} \right)dt} \\&\,\,\,\,\,\,\,\,\, = - \int {\left( {{t^3} - {t^5}} \right)dt = - \left\{ {\frac{{{t^4}}}{4} - \frac{{{t^6}}}{6}} \right\} + C} \\&\,\,\,\,\,\,\,\,\, = - \left\{ {\frac{{{{\cos }^4}x}}{4} - \frac{{{{\cos }^6}x}}{6}} \right\} + C = \frac{{{{\cos }^6}x}}{6} - \frac{{{{\cos }^4}x}}{4} + C\end{align}

## Chapter 7 Ex.7.3 Question 6

Find the integral of $$\sin x\sin 2x\sin 3x$$

### Solution

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

\begin{align}\therefore \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 5x\;dx} } \\&= \frac{1}{4}\left[ {\frac{{ - \cos 2x}}{2}} \right] - \frac{1}{2}\int {\left\{ {\frac{1}{2}\sin \left( {x + 5x} \right) + \frac{1}{2}\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}\left[ {\frac{{ - \cos 6x}}{6} + \frac{{\cos 4x}}{4}} \right] + C\\& = \frac{{ - \cos 2x}}{8} - \frac{1}{8}\left[ {\frac{{ - \cos 6x}}{3} + \frac{{\cos 4x}}{2}} \right] + C\\& = \frac{1}{8}\left[ {\frac{{\cos 6x}}{3} - \frac{{\cos 4x}}{2} - \cos 2x} \right] + C\end{align}

## Chapter 7 Ex.7.3 Question 7

Find the integral of $$\sin 4x\sin 8x$$

### Solution

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

\begin{align}&\therefore \int {\sin 4x\sin 8x\;dx = \int {\left\{ {\frac{1}{2}\cos \left( {4x - 8x} \right) - \frac{1}{2}\cos \left( {4x + 8x} \right)} \right\}dx} } \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int {\left( {\cos \left( { - 4x} \right) - \cos 12x} \right)dx} \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int {\left( {\cos 4x - \cos 12x} \right)dx} \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left[ {\frac{{\sin 4x}}{4} - \frac{{\sin 12x}}{{12}}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{align}

## Chapter 7 Ex.7.3 Question 8

Find the integral of $$\frac{{1 - \cos x}}{{1 + \cos x}}$$

### Solution

\begin{align} \frac{{1 - \cos x}}{{1 + \cos x}} &= \frac{{2\;{{\sin }^2}\frac{x}{2}}}{{2\;{{\cos }^2}\frac{x}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {2\;{{\sin }^2}\frac{x}{2} = 1 - \cos x\,\,{\rm{and\; }}2\;{{\cos }^2}\frac{x}{2} = 1 + \cos x} \right]\\&= {\tan ^2}\frac{x}{2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left( {{{\sec }^2}\frac{x}{2} - 1} \right)\\\therefore \frac{{1 - \cos x}}{{1 + \cos x}}dx &= \int {\left( {{{\sec }^2}\frac{x}{2} - 1} \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left[ {\frac{{\tan \frac{x}{2}}}{{\frac{1}{2}}} - x} \right] + C\\&= 2\tan \frac{x}{2} - x + C\end{align}

## Chapter 7 Ex.7.3 Question 9

Find the integral of $$\frac{{\cos x}}{{1 + \cos x}}$$

### Solution

\begin{align} \frac{{\cos x}}{{1 + \cos x}} &= \frac{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}{{2\;{{\cos }^2}\frac{x}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {as \;\cos x = {{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}\,\,{\rm{and }}\cos x = 2{{\;\cos }^2}\frac{x}{2} - 1} \right]\\&= \frac{1}{2}\left[ {1 - {{\tan }^2}\frac{x}{2}} \right] \\ \\ \therefore \int {\frac{{\cos x}}{{1 + \cos x}}\;dx = \frac{1}{2}\int {\left( {1 - {{\tan }^2}\frac{x}{2}} \right)dx} } \\&= \frac{1}{2}\int {\left( {1 - {{\sec }^2}\frac{x}{2} + 1} \right)dx} \\&= \frac{1}{2}\int {\left( {2 - {{\sec }^2}\frac{x}{2}} \right)dx} \\&= \frac{1}{2}\left[ {2x - \frac{{\tan \frac{x}{2}}}{{\frac{1}{2}}}} \right] + C\\&= x - \tan \frac{x}{2} + C\end{align}

## Chapter 7 Ex.7.3 Question 10

Find the integral of $${\sin ^4}x$$

### Solution

\begin{align}{\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{1}{2}\cos 4x - 2\cos 2x} \right]\\ &= \frac{1}{4}\left[ {\frac{3}{2} + \frac{1}{2}\cos 4x - 2\cos 2x} \right] \\ \therefore \int {{{\sin }^4}x\;dx = \frac{1}{4}\int {\left[ {\frac{3}{2} + \frac{1}{2}\cos 4x - 2\cos 2x} \right]} dx} \\&= \frac{1}{4}\left[ {\frac{3}{2}x + \frac{1}{2}\left( {\frac{{\sin 4x}}{4}} \right) - 2 \times \frac{{\sin 2x}}{2}} \right] + C \\&= \frac{1}{8}\left[ {3x + \frac{{\sin 4x}}{4} - 2\sin 2x} \right] + C \\&= \frac{{3x}}{8} - \frac{1}{4}\sin 2x + \frac{1}{{32}}\sin 4x + C\end{align}

## Chapter 7 Ex.7.3 Question 11

Find the integral of $${\cos ^4}2x$$

### Solution

\begin{align}{\cos ^4}2x &= {\left( {{{\cos }^2}2x} \right)^2}\\ &= {\left( {\frac{{1 + \cos 4x}}{2}} \right)^2}\\& = \frac{1}{4}\left[ {1 + {{\cos }^2}4x + 2\cos 4x} \right]\\ &= \frac{1}{4}\left[ {1 + \left( {\frac{{1 + \cos 8x}}{2}} \right) + 2\cos 4x} \right]\\ &= \frac{1}{4}\left[ {1 + \frac{1}{2} + \frac{{\cos 8x}}{2} + 2\cos 4x} \right]\\ &= \frac{1}{4}\left[ {\frac{3}{2} + \frac{{\cos 8x}}{2} + 2\cos 4x} \right]\end{align}

\begin{align} \therefore \int {{{\cos }^4}2x\;dx = \int {\left( {\frac{3}{8} + \frac{{\cos 8x}}{8} + \frac{{\cos 4x}}{2}} \right)} } dx\\ = \frac{3}{8}x + \frac{1}{{64}}\sin 8x + \frac{1}{8}\sin 4x + C\end{align}

## Chapter 7 Ex.7.3 Question 12

Find the integral of $$\frac{{{{\sin }^2}x}}{{1 + \cos x}}$$

### Solution

\begin{align}\frac{{{{\sin }^2}x}}{{1 + \cos x}}& = \frac{{{{\left( {2\sin \frac{x}{2}\cos \frac{x}{2}} \right)}^2}}}{{2{{\cos }^2}\frac{x}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[as {\;\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2};\cos x = 2{{\cos }^2}\frac{x}{2} - 1} \right]\\& = \frac{{4{{\sin }^2}\frac{x}{2}{{\cos }^2}\frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}\\ &= 2{\sin ^2}\frac{x}{2}\\ &= 1 - \cos x\\ \therefore \int {\frac{{{{\sin }^2}x}}{{1 + \cos x}}dx = \int {\left( {1 - \cos x} \right)dx} } \\&= x - \sin x + C\end{align}

## Chapter 7 Ex.7.3 Question 13

Find the integral of $$\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}$$

### Solution

\begin{align}\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }} &= \frac{{ - 2\sin \frac{{2x + 2\alpha }}{2}\sin \frac{{2x - 2\alpha }}{2}}}{{ - 2\sin \frac{{x + \alpha }}{2}\sin \frac{{x - \alpha }}{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[as {\;\cos C - \cos D = - 2\sin \frac{{C + D}}{2}\sin \frac{{C - D}}{2}} \right]\\&= \frac{{\sin \left( {x + \alpha } \right)\sin \left( {x - \alpha } \right)}}{{\sin \left( {\frac{{x + \alpha }}{2}} \right)\sin \left( {\frac{{x - \alpha }}{2}} \right)}}\\&= \frac{{\left[ {2\sin \left( {\frac{{x + \alpha }}{2}} \right)\cos \left( {\frac{{x - \alpha }}{2}} \right)} \right]\left[ {2\sin \left( {\frac{{x - \alpha }}{2}} \right)\cos \left( {\frac{{x + \alpha }}{2}} \right)} \right]}}{{\sin \left( {\frac{{x + \alpha }}{2}} \right)\sin \left( {\frac{{x - \alpha }}{2}} \right)}}\\&= 4\cos \left( {\frac{{x + \alpha }}{2}} \right)\cos \left( {\frac{{x - \alpha }}{2}} \right)\\&= 2\left[ {\cos \left( {\frac{{x + \alpha }}{2} + \frac{{x - \alpha }}{2}} \right) + \cos \left( {\frac{{x + \alpha }}{2} - \frac{{x - \alpha }}{2}} \right)} \right]\\&= 2\left[ {\cos \left( x \right) + \cos \alpha } \right]\\&= 2\cos x + 2\cos \alpha \\\therefore \int {\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}dx = \int {2\cos x + 2\cos \alpha } } \;dx\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 2\left[ {\sin x + x\cos \alpha } \right] + C\end{align}

## Chapter 7 Ex.7.3 Question 14

Find the integral of $$\frac{{\cos x - \sin x}}{{1 + \sin 2x}}$$

### Solution

\begin{align}\frac{{\cos x - \sin x}}{{1 + \sin 2x}} &= \frac{{\cos x - \sin x}}{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2\sin x\cos x}}\\& \qquad\left[as {{{\;\sin }^2}x + {{\cos }^2}x = 1;\sin 2x = 2\sin x\cos x} \right]\,\,\\& = \frac{{\cos x - \sin x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\end{align}

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

\begin{align}&\therefore \left( {\cos x - \sin x} \right)dx = dt\\ &\Rightarrow \;\int {\frac{{\cos x - \sin x}}{{1 + \sin 2x}}} dx = \int {\frac{{\cos x - \sin x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}dx} \\&= \int {\frac{{dt}}{{{t^2}}}} \\&= \int {{t^{ - 2}}dt} \\&= - {t^{ - 1}} + C\\&= - \frac{1}{t} + C\\&= \frac{{ - 1}}{{\sin x + \cos x}} + C\end{align}

## Chapter 7 Ex.7.3 Question 15

Find the integral of $${\tan ^3}2x\sec 2x$$

### Solution

\begin{align}{\tan ^3}2x\sec 2x &= {\tan ^2}2x\tan 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} - \int {\tan 2x\sec 2x} } \\&= \int {{{\sec }^2}2x\tan 2x\sec 2x - \frac{{\sec 2x}}{2} + C} \\&{\rm{Let }}\sec 2x = t\\&\therefore 2\sec 2x\tan 2xdx = dt\\&\therefore \int {{{\tan }^3}2x\sec 2xdx = \frac{1}{2}\int {{t^2}dt - \frac{{\sec 2x}}{2} + C} } \\& = \frac{{{t^3}}}{6} - \frac{{\sec 2x}}{2} + C\\&= \frac{{{{\left( {\sec 2x} \right)}^3}}}{6} - \frac{{\sec 2x}}{2} + C\end{align}

## Chapter 7 Ex.7.3 Question 16

Find the integral of $${\tan ^4}x$$

### Solution

\begin{align}&{\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}xdx = \int {{{\sec }^2}x{{\tan }^2}xdx - \int {{{\sec }^2}xdx + \int {1dx} } } } \\& = \int {{{\sec }^2}x{{\tan }^2}xdx - \tan x + x + C\,\,\,\,\,\,\,} \ldots \left( 1 \right)\end{align}

Consider $${\sec ^2}x{\tan ^2}xdx$$

Let $$\tan x = t \Rightarrow \;{\sec ^2}xdx = dt$$

$$\Rightarrow \;\int {{{\sec }^2}x{{\tan }^2}xdx = \int {{t^2}dt} = \frac{{{t^3}}}{3} = \frac{{{{\tan }^3}x}}{3}}$$

From equation (1), we get

$$\int {{{\tan }^4}xdx = \frac{1}{3}{{\tan }^3}x - \tan x + x + C}$$

## Chapter 7 Ex.7.3 Question 17

Find the integral of $$\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}$$

### Solution

\begin{align}\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}} &= \frac{{{{\sin }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}} + \frac{{{{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}\\& = \frac{{\sin x}}{{{{\cos }^2}x}} + \frac{{\cos x}}{{{{\sin }^2}x}}\\& = \tan x\sec x + \cot x\;{\rm{cosec}}x\\&\therefore \int {\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx = \int {\left( {\tan x\sec x + \cot x\;{\rm{cosec}}x} \right)dx} } \\&= \sec x - {\rm{cosec}}x + C\end{align}

## Chapter 7 Ex.7.3 Question 18

Find the integral of $$\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}$$

### Solution

\begin{align}&\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}\\ &= \frac{{\cos 2x + \left( {1 - \cos 2x} \right)}}{{{{\cos }^2}x}}\,\,\,\,\,\,\,\,\,\left[ {\cos 2x = 1 - 2{{\sin }^2}x} \right]\\& = \frac{1}{{{{\cos }^2}x}} = {\sec ^2}x\\&\therefore \int {\frac{{\cos 2x + 2{{\sin }^2}x}}{{{{\cos }^2}x}}} dx = \int {{{\sec }^2}xdx = \tan x + C}\end{align}

## Chapter 7 Ex.7.3 Question 19

Find the integral of $$\frac{1}{{\sin x{{\cos }^3}x}}$$

### Solution

\begin{align}\frac{1}{{\sin x{{\cos }^3}x}} &= \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x{{\cos }^3}x}}\\& = \frac{{\sin x}}{{{{\cos }^3}x}} + \frac{1}{{\sin x\cos x}}\\&= \tan x{\sec ^2}x + \frac{{\frac{1}{{{{\cos }^2}x}}}}{{\frac{{\sin x\cos x}}{{{{\cos }^2}x}}}}\\&= \tan x{\sec ^2}x + \frac{{{{\sec }^2}x}}{{\tan x}}\\&\therefore \int {\frac{1}{{\sin x{{\cos }^3}x}}dx = \int {\tan x{{\sec }^2}xdx + \int {\frac{{{{\sec }^2}x}}{{\tan x}}dx} } }\end{align}

Let $$\tan x = t \Rightarrow \;{\sec ^2}xdx = dt$$

\begin{align}& \Rightarrow \;\int {\frac{1}{{\sin x{{\cos }^3}x}}dx = \int {tdt + \int {\frac{1}{t}dt} } } \\ &= \frac{{{t^2}}}{2} + \log \left| t \right| + C\\& = \frac{1}{2}{\tan ^2}x + \log \left| {\tan x} \right| + C\end{align}

## Chapter 7 Ex.7.3 Question 20

Find the integral of $$\frac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}$$

### Solution

\begin{align}\frac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}& = \frac{{\cos 2x}}{{{{\cos }^2}x + \sin {x^2} + 2\cos x\sin x}}= \frac{{\cos 2x}}{{1 + \sin 2x}}\\&\therefore \int {\frac{{\cos 2x}}{{{{\left( {\cos x + \sin x} \right)}^2}}}dx = \int {\frac{{\cos 2x}}{{\left( {1 + \sin 2x} \right)}}dx} }\end{align}

Let $$1 + \sin 2x = t$$

\begin{align} &\Rightarrow \;2\cos 2xdx = dt\\&\therefore \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\\& = \frac{1}{2}\log \left| {1 + \sin 2x} \right| + C\\ &= \frac{1}{2}\log \left| {{{\left( {\sin x + \cos x} \right)}^2}} \right| + C\\ &= \log \left| {\sin x + \cos x} \right| + C\end{align}

## Chapter 7 Ex.7.3 Question 21

Find the integral of $${\sin ^{ - 1}}\left( {\cos x} \right)$$

### Solution

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

Let $$\cos x = t$$

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

\begin{align} \Rightarrow \;\left( { - \sin x} \right)dx &= dt\\dx &= \frac{{ - dt}}{{\sin x}}\\dx &= \frac{{ - dt}}{{\sqrt {1 - {t^2}} }}\\&\therefore \int {{{\sin }^{ - 1}}\left( {\cos x} \right)dx = \int {{{\sin }^{ - 1}}t\left( {\frac{{ - dt}}{{\sqrt {1 - {t^2}} }}} \right)} } \\ &= - \int {\frac{{{{\sin }^{ - 1}}t}}{{\sqrt {1 - {t^2}} }}}\end{align}

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

\begin{align} &\Rightarrow \;\frac{1}{{\sqrt {1 - {t^2}} }}dt = du\\&\therefore \int {{{\sin }^{ - 1}}\left( {\cos x} \right)dx = - \int {udu} } \\& = - \frac{{{u^2}}}{2} + C\\& = \frac{{ - {{\left( {{{\sin }^{ - 1}}t} \right)}^2}}}{2} + C\\& = \frac{{ - {{\left[ {{{\sin }^{ - 1}}\left( {\cos x} \right)} \right]}^2}}}{2} + C\,\,\,\,\, \ldots \left( 1 \right)\end{align}

We know that,

\begin{align}&{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}\\&\therefore {\sin ^{ - 1}}\left( {\cos x} \right) = \frac{\pi }{2} - {\cos ^{ - 1}}\left( {\cos x} \right) = \left( {\frac{\pi }{2} - x} \right)\end{align}

Substituting in equation (1), we get

\begin{align}&\int {{{\sin }^{ - 1}}\left( {\cos x} \right)dx = \frac{{ - {{\left[ {\frac{\pi }{2} - x} \right]}^2}}}{2} + C} \\& = - \frac{1}{2}\left( {\frac{{{\pi ^2}}}{4} + {x^2} - \pi x} \right) + C\\ &= - \frac{{{\pi ^2}}}{8} - \frac{{{x^2}}}{2} + \frac{{\pi x}}{2} + C\\ &= \frac{{\pi x}}{2} - \frac{{{x^2}}}{2} + \left( {C - \frac{{{\pi ^2}}}{8}} \right)\\ &= \frac{{\pi x}}{2} - \frac{{{x^2}}}{2} + {C_1}\end{align}

## Chapter 7 Ex.7.3 Question 22

Find the integral of $$\frac{1}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}$$

### Solution

\begin{align}\frac{1}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}} &= \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\frac{{\sin \left( {a - b} \right)}}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}} \right]\\ &= \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\frac{{\sin \left[ {\left( {x - b} \right) - \left( {x - a} \right)} \right]}}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}} \right]\\& = \frac{1}{{\sin \left( {a - b} \right)}}\frac{{\left[ {\sin \left( {x - b} \right)\cos \left( {x - a} \right) - \cos \left( {x - b} \right)\sin \left( {x - a} \right)} \right]}}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}\\& = \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\tan \left( {x - b} \right) - \tan \left( {x - a} \right)} \right]\\& \Rightarrow \;\int {\frac{1}{{\cos \left( {x - a} \right)\cos \left( {x - b} \right)}}dx = \frac{1}{{\sin \left( {a - b} \right)}}\int {\left[ {\tan \left( {x - b} \right) - \tan \left( {x - a} \right)} \right]} dx} \\ &= \frac{1}{{\sin \left( {a - b} \right)}}\left[ { - \log \left| {\cos \left( {x - b} \right)} \right| + \log \left| {\cos \left( {x - a} \right)} \right|} \right]\\ &= \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\log \left| {\frac{{\cos \left( {x - a} \right)}}{{\cos \left( {x - b} \right)}}} \right|} \right] + C\end{align}

## Chapter 7 Ex.7.3 Question 23

$$\int {\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx}$$ is equal to

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

(B) $$\tan x + \cos ecx + C$$

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

(D) $$\tan x + \sec x + C$$

### Solution

\begin{align}&\int {\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}dx = \int {\left( {\frac{{{{\sin }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}} - \frac{{{{\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\end{align}

Thus, the correct option is A.

## Chapter 7 Ex.7.3 Question 24

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

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

(B) $$\tan \left( {x{e^x}} \right) + C$$

(C) $$\tan \left( {{e^x}} \right) + C$$

(D) $$\cot \left( {{e^x}} \right) + C$$

### Solution

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

Put, $${e^x}x = t$$

\begin{align} &\Rightarrow \;\left( {{e^x}x + {e^x}.1} \right)dx = dt\\&{e^x}\left( {x + 1} \right)dx = dt\\&\therefore \int {\frac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {{e^x}x} \right)}}} dx = \int {\frac{{dt}}{{{{\cos }^2}t}}} \\& = \int {{{\sec }^2}tdt} \\& = \tan t + C\\ &= \tan \left( {{e^x}x} \right) + C\end{align}

Thus, the correct answer is $$B.$$

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