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

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

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
$\\\cos (2x+1) =t\\ \sin (2x+1)dx = -dt/2$

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

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:

$I = \int \sin^3x.\cos^3x\ dx$

rewrite the integral as follows

$\\=\int cos^3x.sin^2x.\sin x\ dx\\ =\int cos^3x(1-\cos^2x)\sin x\ dx$
Let $\cos x = t \Rightarrow \sin x dx =-dt$

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

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

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

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

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

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

Answer:

We know the identities

$\\1+\cos 2A = 2\cos^2A\\ 1-\cos 2A = 2\sin^2 A$

Using the above relations we can write

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

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

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

Answer:

The integral is rewritten using trigonometric identities

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

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

Answer:

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

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

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

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

Answer:

$cos^42x=cos^32xcos2x$

now using the identity

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

$cos^32xcos2x=\frac{cos6x +3cos2x}{4}cos2x=\frac{cos6xcos2x+3cos^22x}{4}$

now using the below two identities

$cosacosb=\frac{cos(a+b)+cos(a-b)}{2}\\and\ cos^22x=\frac{1+cos4x}{2}\\$

the value

$cos^42x=cos^32xcos2x\\=\frac{cos6xcos2x+3cos^22x}{4}=\frac{cos 4x+cos8x}{8}+\frac{3}{4}\frac{1+cos4x}{2}$ .

the integral of the given function can be written as

$\int cos^42x=\int \frac{cos 4x+cos8x}{8}+\int \frac{3}{4}\frac{1+cos4x}{2}\\ \\=\frac{3}{8}x+\frac{sin4x}{8}+\frac{sin8x}{64}+C$

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

Answer:

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

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

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

$\therefore \int \frac{sin^22x}{1+\cos x} = \int (1-\cos x)dx$
$\\= \int 1dx-\int\cos x\ dx\\ =x-\sin x+C$

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

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

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

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

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

Answer:

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

$\\sin x+\cos x =t\\ \therefore (\cos x-\sin x)dx = dt$

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

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

Answer:

$\tan^32x.\sec 2x = \tan^22x.\tan 2x.\sec 2x$
$\\= (\sec^22x-1).\tan 2x.\sec 2x\\ =\sec^22x.\tan 2x-\tan 2x.\sec 2x$

Therefore integration of $\tan ^ 3 2x \sec 2x$ =
$\\=\int\sec^22x.\tan 2x\ dx-\int\tan 2x.\sec 2x\ dx\\ =\int\sec^22x.\tan 2x\ dx-\sec 2x/2+C\\$ .....................(i)
Let assume

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

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

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

Answer:

the given question can be rearranged using trigonometric identities

$tan^4x=(\sec^2x-1).\tan^2x\\ =\sec^2x.\tan^2x-\tan^2x\\ =\sec^2x.\tan^2x-\sec^2x+1$

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

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

now the final solution is,

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

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

Answer:

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

now splitting the terms we can write

$\\=\frac{\sin^3x}{\sin^2x.\cos^2x}+\frac{\cos^3x}{\sin^2x.\cos^2x}\\ =\frac{\sin x}{cos^2x}+\frac{\cos x}{\sin^2x}\\ =\tan x.\sec x+\cot xcosec x$

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

$\\=\int (\tan x\sec x+\cot xcosec x)dx\\ =\sec x-cosec\ x+C$

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

Answer:

The integral of the above equation is

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

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

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

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

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

Answer:

we know that $cos2x= cos^2x-sin^2x$
therefore,

$\frac{\cos 2x }{( \cos x + \sin x )^2}$
$\frac{\cos 2x}{1+\sin 2x}\\ \Rightarrow \int \frac{\cos 2x}{1+\sin 2x}\\$ let $1+sin2x =t \Rightarrow 2cos2x\ dx = dt$
Now the given integral can be written as

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

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

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

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

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

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

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

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

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

Question 23: Choose the correct answer

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

Answer:

The correct option is (A)

On reducing the above integral becomes $\sec^2x-csc^2x$
$\int\sec^2x-csc^2x\ dx = \tan x+ \cot x+C$

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

$\\(A) -\cot (ex^x) + C \\\\ (B) \tan (xe^x) + C\\\\ (C) \tan (e^x) + C \\\\ (D) \cot (e^x) + C$

Answer:

The correct option is (B)

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

therefore,

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