NCERT Solutions for Exercise 7.3 Class 12 Maths Chapter 7 - Integrals

Question 2: Find the integrals of the functions $\sin 3x\cos 4x$

Answer:

Using identity $\sin A\cos B=1/2[sin(A+B)+sin(A-B)]$

, therefore the given integral can be written as

$\int \sin 3x\cos 4x=\frac{1}{2}\int \sin (3x+4x)+\sin (3x-4x)dx$
$=\frac{1}{2}\int \sin (7x)-\sin (x)dx$
$=\frac{1}{2}[\int \sin (7x)dx-\int \sin xdx]$
$=\frac{1}{2}[\frac{-\cos 7x}{7}+\cos x]+C$
$=\frac{\cos x}{2}-\frac{\cos 7x}{14}+C$

Question 3: Find the integrals of the functions $\cos 2x\cos 4x\cos 6x$

Answer:

Using identity
$cosAcosB=\frac{1}{2}[cos(A+B)+cos(A-B)]$

$\int \cos 2x.\cos 4x.\cos 6x=\int \cos 2x.\frac{1}{2}[(\cos 10x)+\cos 2x]dx$

Again use the same identity mentioned in the first line

$=\frac{1}{2}\int (\cos 2x\cdot \cos 10x+\cos 2x\cdot \cos 2x)dx$
$=\frac{1}{2}\int \frac{1}{2}(\cos 12x+\cos 8x)dx+\frac{1}{2}\int \left(\frac{1+\cos 4x}{2}\right)dx$
$=\frac{\sin 12x}{48}+\frac{\sin 8x}{32}+\frac{\sin 4x}{16}+\frac{x}{4}+C$

Question 4: Find the integrals of the functions ${\sin }^3(2x+1)$

Answer:

$\int {\sin }^3(2x+1)dx=\int {\sin }^2(2x+1).\sin (2x+1)dx$

The integral can be written as

$=\int (1-{\cos }^2(2x+1)).\sin (2x+1)dx$
Let
$$\begin{gathered} \cos (2x+1)=t \\ \sin (2x+1)dx=-dt/2 \end{gathered}$$

$$\begin{gathered} =\frac{-1}{2}\int (1-t^2)dt \\ =\frac{-1}{2}[t-t^3/3] \\ =\frac{t^3}{6}-\frac{t}{2} \end{gathered}$$

Now, replace the value of t, we get;

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

Question 5: Find the integrals of the functions ${\sin }^{3}x{\cos }^{3}x$

Answer:

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

rewrite the integral as follows

$$\begin{gathered} =\int co{s}^{3}x.si{n}^{2}x.\sin xdx \\ =\int co{s}^{3}x(1-{\cos }^{2}x)\sin xdx \end{gathered}$$
Let $\cos x=t\Rightarrow \sin xdx=-dt$

$$\begin{gathered} =-\int t^3(1-t^2)dt \\ =-\int (t^3-t^5)dt \\ =-[\frac{t^4}{4}]+[\frac{t^6}{6}]+C \\ =\frac{{\cos }^{6}x}{6}-\frac{co{s}^{4}x}{4}+C \end{gathered}$$ ......(replace the value of t as $cosx$ )

Question 6: Find the integrals of the functions $\sin x\sin 2x\sin 3x$

Answer:

Using the formula
$sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

we can write the integral as follows

$\int \sin x\cdot \sin 2x\cdot \sin 3xdx=\int \sin x\cdot \frac{1}{2}[\cos x-\cos 5x]dx$
$=\frac{1}{2}\int [\sin x\cdot \cos x-\sin x\cdot \cos 5x]dx$
$=\frac{1}{2}\int \frac{\sin 2x}{2}dx-\frac{1}{2}\int \sin x\cdot \cos 5xdx$
$=-\frac{\cos 2x}{8}-\frac{1}{4}\int [\sin 6x-\sin 4x]dx$
$=-\frac{\cos 2x}{8}-\frac{1}{4}[\frac{-\cos 6x}{6}+\frac{\cos 4x}{4}]$
$=-\frac{\cos 2x}{8}+\frac{\cos 6x}{24}-\frac{\cos 4x}{16}+C$

Question 7: Find the integrals of the functions $\sin 4x\sin 8x$

Answer:

Using identity

$sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

we can write the following integral as

$\sin 4x\sin 8x$ =
$$\begin{gathered} =\frac{1}{2}\int (\cos 4x-\cos 12x)dx \\ =\frac{1}{2}[\int \cos 4xdx-\int \cos 12xdx] \\ =\frac{\sin 4x}{8}-\frac{\sin 12x}{24}+C \end{gathered}$$

Question 8: Find the integrals of the functions $\frac{1-\cos x}{1+\cos x}$

Answer:

We know the identities

$$\begin{gathered} 1+\cos 2A=2{\cos }^{2}A \\ 1-\cos 2A=2{\sin }^{2}A \end{gathered}$$

Using the above relations we can write

$\frac{1-\cos x}{1+\cos x}=\frac{{\sin }^{2}x/2}{{\cos }^{2}x/2}={\tan }^{2}x/2$

$=\int {\tan }^{2}x/2=\int ({\sec }^{2}x/2-1)dx$
$$\begin{gathered} =\int ({\sec }^{2}x/2)dx-\int dx \\ =2[\tan x/2]-x+C \end{gathered}$$

Question 9: Find the integrals of the functions $\frac{\cos x}{1+\cos x}$

Answer:

The integral is rewritten using trigonometric identities

$\frac{\cos x}{1+\cos x}=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}=\frac{1}{2}[1-{\tan }^2\frac{x}{2}]$
$=\int \frac{1}{2}[1-{\tan }^2\frac{x}{2}]dx$
$=\frac{1}{2}\int [1-({\sec }^2\frac{x}{2}-1)]dx$
$=\frac{1}{2}\int [2-{\sec }^2\frac{x}{2}]dx$
$=x-\tan \frac{x}{2}+C$

Question 10: Find the integrals of the functions ${\sin }^{4}x$

Answer:

${\sin }^{4}x$ can be written as follows using trigonometric identities

$={\sin }^{2}x\cdot {\sin }^{2}x$
$=\frac{1}{4}(1-\cos 2x{)}^2$
$=\frac{1}{4}(1+{\cos }^{2}2x-2\cos 2x)$
$=\frac{1}{4}(1+\frac{1}{2}(1+\cos 4x)-2\cos 2x)$
$=\frac{3}{8}+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}$

Therefore,
$\Rightarrow \int {\sin }^{4}xdx=\int \frac{3}{8}dx+\frac{1}{8}\int \cos 4xdx-\frac{1}{2}\int \cos 2xdx$
$=\frac{3x}{8}+\frac{\sin 4x}{32}-\frac{\sin 2x}{4}+C$

Question 11: Find the integrals of the functions ${\cos }^{4}2x$

Answer:

$co{s}^{4}2x=co{s}^{3}2xcos2x$

now using the identity

$co{s}^{3}x=\frac{cos3x+3cosx}{4}$

$co{s}^{3}2xcos2x=\frac{cos6x+3cos2x}{4}cos2x=\frac{cos6xcos2x+3co{s}^{2}2x}{4}$

now using the below two identities

$$\begin{gathered} cosacosb=\frac{cos(a+b)+cos(a-b)}{2} \\ andco{s}^{2}2x=\frac{1+cos4x}{2} \end{gathered}$$

the value

$$\begin{gathered} co{s}^{4}2x=co{s}^{3}2xcos2x \\ =\frac{cos6xcos2x+3co{s}^{2}2x}{4}=\frac{cos4x+cos8x}{8}+\frac{3}{4}\frac{1+cos4x}{2} \end{gathered}$$ .

the integral of the given function can be written as

$$\begin{gathered} \int co{s}^{4}2x=\int \frac{cos4x+cos8x}{8}+\int \frac{3}{4}\frac{1+cos4x}{2} \\ =\frac{3}{8}x+\frac{sin4x}{8}+\frac{sin8x}{64}+C \end{gathered}$$

Question 12: Find the integrals of the functions $\frac{{\sin }^{2}x}{1+\cos x}$

Answer:

Using trigonometric identities we can write the given integral as follows.

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

$$\begin{gathered} =\frac{4{\sin }^2\frac{x}{2}{\cos }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}} \\ =2{\sin }^2\frac{x}{2} \\ =1-\cos x \end{gathered}$$

$\therefore \int \frac{si{n}^{2}2x}{1+\cos x}=\int (1-\cos x)dx$
$$\begin{gathered} =\int 1dx-\int \cos xdx \\ =x-\sin x+C \end{gathered}$$

Question 13: Find the integrals of the functions $\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }$

Answer:

We know that,

$\cos A-\cos B=-2\sin (\frac{A+B}{2})\sin (\frac{A-B}{2})$

Using this identity we can rewrite the given integral as

$\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\frac{-2\sin \frac{2x+2\alpha }{2}\sin \frac{2x-2\alpha }{2}}{-2\sin \frac{x+\alpha }{2}\sin \frac{x-\alpha }{2}}$

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

$\therefore \int \frac{\cos 2x-\cos 2\alpha }{\cos x-\cos \alpha }=\int 2\cos xdx+\int 2\cos \alpha dx$
$=2[\sin x+x\cos \alpha ]+C$

Question 14: Find the integrals of the functions $\frac{\cos x-\sin x}{1+\sin 2x}$

Answer:

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

$$\begin{gathered} sinx+\cos x=t \\ \therefore (\cos x-\sin x)dx=dt \end{gathered}$$

Now,
$$\begin{gathered} =\int \frac{dt}{t^2} \\ =\int t^{-}2dt \\ =-t^{-1}+C \\ =-\frac{1}{(\sin x+\cos x)}+C \end{gathered}$$

Question 15: Find the integrals of the functions ${\tan }^{3}2x\sec 2x$

Answer:

${\tan }^{3}2x.\sec 2x={\tan }^{2}2x.\tan 2x.\sec 2x$
$$\begin{gathered} =({\sec }^{2}2x-1).\tan 2x.\sec 2x \\ ={\sec }^{2}2x.\tan 2x-\tan 2x.\sec 2x \end{gathered}$$

Therefore integration of ${\tan }^{3}2x\sec 2x$ =
$$\begin{gathered} =\int {\sec }^{2}2x.\tan 2xdx-\int \tan 2x.\sec 2xdx \\ =\int {\sec }^{2}2x.\tan 2xdx-\sec 2x/2+C \end{gathered}$$ .....................(i)
Let assume

$\sec 2x=t$
So, that $2\sec 2x.\tan 2xdx=dt$
Now, the equation (i) becomes,

$$\begin{gathered} \Rightarrow \frac{1}{2}\int t^{2}dt-\frac{\sec 2x}{2}+C \\ \Rightarrow \frac{t^3}{6}-\frac{\sec 2x}{2}+C \\ =\frac{(\sec 2x{)}^3}{6}-\frac{\sec 2x}{2}+C \end{gathered}$$

Question 16: Find the integrals of the functions ${\tan }^{4}x$

Answer:

the given question can be rearranged using trigonometric identities

$$\begin{gathered} ta{n}^{4}x=({\sec }^{2}x-1).{\tan }^{2}x \\ ={\sec }^{2}x.{\tan }^{2}x-{\tan }^{2}x \\ ={\sec }^{2}x.{\tan }^{2}x-{\sec }^{2}x+1 \end{gathered}$$

Therefore, the integration of ${\tan }^{4}x$ = $$\begin{gathered} =\int {\sec }^{2}x.{\tan }^{2}xdx-\int {\sec }^{2}xdx+\int dx \\ =(\int {\sec }^{2}x.{\tan }^{2}xdx)-\tan x+x+C \end{gathered}$$ ...................(i)
Considering only $\int {\sec }^{2}x.{\tan }^{2}xdx$
let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

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

now the final solution is,

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

Question 17: Find the integrals of the functions $\frac{{\sin }^{3}x+{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}$

Answer:

$\frac{{\sin }^{3}x+{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}$

now splitting the terms we can write

$$\begin{gathered} =\frac{{\sin }^{3}x}{{\sin }^{2}x.{\cos }^{2}x}+\frac{{\cos }^{3}x}{{\sin }^{2}x.{\cos }^{2}x} \\ =\frac{\sin x}{co{s}^{2}x}+\frac{\cos x}{{\sin }^{2}x} \\ =\tan x.\sec x+\cot xcosecx \end{gathered}$$

Therefore, the integration of
$\frac{{\sin }^{3}x+{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}$

$$\begin{gathered} =\int (\tan x\sec x+\cot xcosecx)dx \\ =\sec x-cosecx+C \end{gathered}$$

Question 18: Find the integrals of the functions $\frac{\cos 2x+2{\sin }^{2}x}{{\cos }^{2}x}$

Answer:

The integral of the above equation is

$$\begin{gathered} =\int (\frac{\cos 2x+2{\sin }^{2}x}{{\cos }^{2}x})dx \\ =\int (\frac{\cos 2x+(1-\cos 2x)}{{\cos }^{2}x} \\ =\int \frac{1}{{\cos }^{2}x} \\ =\int {\sec }^{2}xdx=\tan x+C \end{gathered}$$

Thus after evaluation, the value of integral is tanx+ c

Question 19: Find the integrals of the functions $\frac{1}{\sin x{\cos }^{3}x}$

Answer:

Let
We can write 1 = ${\sin }^{2}x+{\cos }^{2}x$
Then, the equation can be written as
$I=\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x{\cos }^{3}x}$

$I=\int (\tan x+\frac{1}{\tan x}){\sec }^{2}xdx$
put the value of tan $x$ = t
So, that ${\sec }^{2}xdx=dt$

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

Question 20: Find the integrals of the functions $\frac{\cos 2x}{(\cos x+\sin x{)}^2}$

Answer:

we know that $cos2x=co{s}^{2}x-si{n}^{2}x$
therefore,

$\frac{\cos 2x}{(\cos x+\sin x{)}^2}$
$$\begin{gathered} \frac{\cos 2x}{1+\sin 2x} \\ \Rightarrow \int \frac{\cos 2x}{1+\sin 2x} \end{gathered}$$ let $1+sin2x=t\Rightarrow 2cos2xdx=dt$
Now the given integral can be written as

$\therefore \int \frac{\cos 2x}{(\cos x+\sin x{)}^2}=\frac{1}{2}\int \frac{1}{t}dt$
$$\begin{gathered} \Rightarrow \frac{1}{2}\log \left|t\right|+C \\ \Rightarrow \frac{1}{2}\log |1+\sin 2x|+C \\ =log|si{n}^{2}x+co{s}^{2}x+2sinxcosx|+C \\ =\frac{1}{2}log|(sinx+cosx{)}^2|+C=log|sinx+cosx|+C \end{gathered}$$

Question 21: Find the integrals of the functions ${\sin }^{-1}(\cos x)$

Answer:

using the trigonometric identities we can evaluate the following integral as follows

$100\int {\sin }^{-1}(\cos x)dx=\int {\sin }^{-1}(\sin (\frac{\pi }{2}-x))dx$
$=\int (\frac{\pi }{2}-x)dx=\frac{\pi x}{2}-\frac{x^2}{2}+C$
$\frac{\cos x-\sin x}{1+2\sin x}=\frac{\cos x-\sin x}{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}$
$=\frac{\cos x-\sin x}{(\sin x+\cos x{)}^2}$

Question 22: Find the integrals of the functions $\frac{1}{\cos (x-a)\cos (x-b)}$

Answer:

Using the trigonometric identities following integrals can be simplified as follows

$\frac{1}{\cos (x-a)\cos (x-b)}=\frac{1}{\sin (a-b)}[\frac{\sin (a-b)}{\cos (x-a)\cos (x-b)}]$

$=\frac{1}{\sin (a-b)}[\frac{\sin [(x-b)-(x-a)]}{\cos (x-a)\cos (x-b)}]$

$=\frac{1}{\sin (a-b)}[\frac{\sin (x-b)\cos (x-a)-\cos (x-b)\sin (x-a)}{\cos (x-a)\cos (x-b)}]$

$=\frac{tan(x-b)-\tan (x-a)}{\sin (a-b)}$

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

Question 23: Choose the correct answer

$\int \frac{{\sin }^{2}x-{\cos }^{2}xdx}{{\sin }^{2}x{\cos }^{2}x}dxisequalto$

Answer:

The correct option is (A)

On reducing the above integral becomes ${\sec }^{2}x-cs{c}^{2}x$
$\int {\sec }^{2}x-cs{c}^{2}xdx=\tan x+\cot x+C$

Question 24: Choose the correct answer $\int \frac{e^x(1+x)}{{\cos }^2(e^{x}x)}dxequals$

$$\begin{gathered} (A)-\cot (e{x}^x)+C \\ (B)\tan (x{e}^x)+C \\ (C)\tan (e^x)+C \\ (D)\cot (e^x)+C \end{gathered}$$

Answer:

The correct option is (B)

Let $e^{x}x=t$ .
So, $(e^x.x+1.e^x)dx=dt$
(1+ $x$ ) $e^{x}dx=dt$

therefore,

$\int \frac{e^x(1+x)}{{\cos }^2(e^x.x)}dx=\int \frac{dt}{{\cos }^{2}t}$
$$\begin{gathered} =\int {\sec }^{2}tdt \\ =\tan t+C \\ =\tan (e^x.x)+C \end{gathered}$$