KC Sinha: Exercise 19.4- Mathematics Solution Class 12 Chapter 19 Indefinite Integrals

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Question 1

$\displaystyle\int sin^{3}xdx$

Sol :

Let z=cosx then dz=-sinxdx

=$\displaystyle\int sin^{3}xdx$

=$\displaystyle\int sin^{2}x.sinxdx$

=$\displaystyle\int (1-cos^{2}x).sinxdx$

=$\displaystyle\int (1-z^{2}).(-dz)$

=$-\displaystyle\int (1-z^{2}).(dz)$

=$-\displaystyle\int (dz)+\displaystyle\int z^{2}dz$

=$-z+\dfrac{z^3}{3}$+c

=$-cosx+\dfrac{cos^{3}x}{3}$+c

Question 2
$\displaystyle\int cos^{5}dx$
Sol :
Let z=sinx then dz=cosxdx

Now, $\displaystyle\int cos^{5}dx$

=$\displaystyle\int cos^{4}x.cosxdx$

=$\displaystyle\int (cos^{2}x)^2.cosxdx$

=$\displaystyle\int (1-sin^{2}x)^2.cosxdx$

=$\displaystyle\int (1+sin^{4}x-2sin^2x).cosxdx$

=$\displaystyle\int (1+z^{4}-2z^2)dz$

=$\displaystyle\int dz+\displaystyle\int z^4dz-2\displaystyle\int z^2dz$

=$z+\dfrac{z^5}{5}-2\dfrac{z^3}{3}$+c

=$sinx+\dfrac{sin^{5}x}{5}-\dfrac{2}{5}sin^{3}x$+c

=$sinx-\dfrac{2}{5}sin^{3}x+\dfrac{sin^{5}x}{5}$+c

Question 3
$\displaystyle\int sin^{2}x.cosxdx$
Sol :
Let z=sinx then dz=cosxdx

=$\displaystyle\int z^2dz$

=$\dfrac{z^3}{3}$+c

=$\dfrac{sin^{3}x}{3}$+c

Question 4
$\displaystyle\int sin^{2}x.cos^{3}xdx$
Sol :
Let z=sinx then dz=cosxdx

Now, $\displaystyle\int z^2(1-z^2)dz$

=$\displaystyle\int z^2-z^4dz$

=$\dfrac{z^3}{3}-\dfrac{z^5}{5}$+c

=$\dfrac{sin^{3}x}{3}-\dfrac{sin^{5}x}{5}$+c

Question 5
$\displaystyle\int cos^{2}x.sin^{3}xdx$
Sol :
Let z=cosx then dz=-sinxdx

Now, $\displaystyle\int cos^{2}x.sin^{3}xdx$

=$cos^2x.sin^2x.sinxdx$

=$\displaystyle\int cos^2x.(1-cos^2x).sinxdx$

=$\displaystyle\int z^2.(1-z^2).(-dz)$

=$-\displaystyle\int z^2-z^4dz$

=$-\dfrac{z^3}{3}+\dfrac{z^5}{5}$+c

=$-\dfrac{cos^3x}{3}+\dfrac{cos^5x}{5}$+c

=$\dfrac{cos^5x}{5}-\dfrac{cos^3x}{3}$+c

Question 6
$\displaystyle\int sin^{\textstyle\frac{3}{4}}x.cosxdx$
Sol :
Let z=sinx then dz=cosxdx

Now, $\displaystyle\int sin^{\textstyle\frac{3}{4}}x.cosxdx$

=$\displaystyle\int z^{\textstyle\frac{3}{4}}.dz$

=$\dfrac{z^{\textstyle\frac{3}{4}+1}}{\frac{3}{4}+1}$+c

=$\dfrac{z^{\textstyle\frac{7}{4}}}{\frac{7}{4}}$+c

=$\dfrac{4}{7}.z^{1/4}$+c

=$\dfrac{4}{7}sin^{\textstyle\frac{7}{4}}x$+c

Question 7
$\int \sin ^{4} x \cdot \cos ^{3} x d x$
Sol :
Let z=sinx then dz=cosxdx
Now , $\int \sin ^{4} x \cdot \cos ^{3} x d x=\int \sin ^{4} x \cdot \cos ^{2} x \cdot \cos x d x$

$\int \sin ^{4} x\left(1-\sin ^{2} x\right) \cdot \cos x d x$

$=\int z^{4}\left(1-z^{2}\right) d z$

$=\int z^{4}-z^{6} d z$

$=\dfrac{2^{5}}{5}-\dfrac{z^{7}}{7}+c$

$=\dfrac{\sin ^{5} x}{5}-\dfrac{\sin ^{7} x}{7}+c$

Question 8
$\int \sin ^{2} x \cdot \cos ^{5} x d x$
Sol :
Let z=sinx then dz=cosxdx
Now , $\int \sin ^{2} x \cdot \cos ^{5} x d x=\int \sin ^{2} x \cdot \cos ^{4} x \cdot \cos x d x$

$=\int \sin ^{2} x\left(\cos ^{2} x\right)^{2} \cdot \cos x d x$

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

$=\int z^{2}\left(1-z^{2}\right)^{2} d z$

$=\int z^{2}\left(1+z^{4}-2z^{2}\right) d z$

$=\int z^{2} d z+\int z^{6} d z-2 \int z^4d z$

$=\dfrac{z^{3}}{3}+\dfrac{z^{7}}{7}-\dfrac{2z^{5}}{5}+c$

$=\dfrac{\sin ^{3} x}{3}+\dfrac{\sin ^{7} x}{7} -\dfrac{2 \sin ^{5} x}{5}+c$

Question 9
$\int \sin ^{3} x \cdot \sin 2 x d x$
Sol :
Let z=sinx then dz=cosxdx

Now , $\int \sin ^{3} x \cdot \sin 2 x d x=\int \sin ^{3} x \cdot 2 \sin x \cdot \cos x d x$

$=2 \int \sin ^{4} x \cdot \cos x d x$

$=2 \int 2^{4} d z$

$=2 \cdot \dfrac{z^{5}}{5}+c=\dfrac{2}{5} \cdot \sin^5 x+c$

Question 10
$\int \sin x \cdot \cot ^{3} x d x$
Sol :
$\int \sin x \cdot \cot ^{3} x d x$

$=\int \sin x \cdot \frac{\cos ^{3} x}{\sin ^{3} x} d x$

$=\int \dfrac{\cos ^{3} x}{\sin ^{2} x} d x$

$=\int \frac{\cos ^{2} x \cdot \cos x d x}{\sin ^{2} x}$

Let z=sinx then dz=cosxdx

Now , $\int \sin x \cdot \cot ^{3} x d x=\int \frac{\cos ^{2} x \cdot \cos x}{\sin ^{2} x} d x$

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

$=\int \frac{\left(1-z^{2}\right)}{z^{2}} dz=\int \frac{1}{z^{2}} d 2-\int \frac{z^{2}}{z^{2}} d 2$

$=\int z^{-2} d 2-\int d z$

$=\dfrac{z^{-2+1}}{-2+1}-z+c$

$=-z^{-1}-z+c$

$=-\dfrac{1}{z}-z+c$

$=-\dfrac{1}{\sin x}-\sin x+c$

=-cosecx-sinx+c

Question 11
$\int \cos x \cdot \tan ^{3} x d x$
Sol :
$=\int \cos x \cdot \frac{\sin ^{3} x}{\cos ^{3} x} d x$

$=\int \frac{\sin ^{2} x \cdot \sin x d x}{\cos ^{2} x}$

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

Let z=cosx then dz=-sinxdx
=-dz=sinxdx

Now , $\int \cos x \cdot \tan ^{3} x d x=\int \frac{\left(1-\cos ^{2} x\right) \cdot \sin x d x}{\cos ^{2} x}$

$=\int \frac{\left(1-z^{2}\right)}{z^{2}}(-d z)$

$=-\int \frac{\left(1-z^{2}\right)}{z^{2}} d z$

$=-\int \frac{1}{z^{2}} d z+\int \frac{z^{2}}{z^{2}} d z$

$=-\int z^{-2} d z+\int d z$

$=-\frac{z^{-2+1}}{-2+1}+z+c$

$=z^{-1}+z+c$

$=\dfrac{1}{z}+z+c$

$=\dfrac{1}{\cos x}+\cos x+c$

$=\dfrac{1}{\cos x}+\cos x+c$

=secx+cosx+c

Question 12
$\int \frac{\cos ^{3} x}{\sqrt{\sin x}} d x$
Sol :
Let z=sinx then dz=cosxdx

Now , $\int \frac{\cos ^{2} x \cdot \cos x d x}{\sqrt{\sin x}}=\int \frac{\left(1-\sin ^{2} x\right) \cos x d x}{\sqrt{\sin x}}$

$=\int \frac{\left(1-z^{2}\right)}{\sqrt{z}} d z=\int \frac{1}{\sqrt{z}} d z-\int \frac{z^{2}}{\sqrt{z}} d z$

$=\int z^{-\frac{1}{2}} d z-\int z^{2-\frac{1}{2}} d z$

$=\int z^{-\frac{1}{2}} d z-\int z^{\frac{3}{2}} d z$

$=\frac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}-\frac{z^{\frac{3}{2}+1}}{\frac{3}{2}+1}+c$

$=2 \cdot z^{\frac{1}{2}}-\frac{2}{5} \cdot z^{\frac{5}{2}}+c$

$=2 \sqrt{2}-\frac{2}{5} z^{\frac{5}{2}}+c$

$=2 \sqrt{2}\left(1-\frac{z^{2}}{5}\right)+c$

$=2 \sqrt{\sin x}\left(1-\frac{\sin ^{2} x}{5}\right)+c$

(ii) $\int \frac{\sec \theta}{\tan ^{2} \theta} d \theta$
Sol :
$\int \frac{\sec \theta}{\tan ^{2} \theta} d \theta=\int \frac{1}{\cos \theta \cdot \frac{\sin ^{2} \theta}{\cos ^{2} \theta}} d \theta$

$=\int \frac{\cos \theta}{\sin ^{2} \theta} d \theta$

Let z=sinθ then dz=cosθdθ

Now  ,  $\int \frac{\sec \theta}{\tan ^{2} \theta} d \theta=\int \frac{\cos \theta}{\sin ^{2} \theta} d \theta$

$=\int \frac{d z}{z^{2}}=\int z^{-2} d z$

$=\frac{z^{-2+1}}{-2+1}+c=-z^{-1}+c$

$=-\frac{1}{2}+c$

$=-\frac{1}{\sin \theta}+c$

=-cosecθ+c

Question 13
$\int \cos ^{2} x \cdot \sin 3 x d x$
Sol :
$=\int \cos ^{2} x \cdot\left(3 \sin x-4 \sin ^{3} x\right) d x$

$=\int \cos ^{2} x \cdot \sin x\left(3-4 \sin ^{2} x\right) d x$

$=\int \cos ^{2} x\left(3-4 \sin ^{2} x\right) \cdot \sin x d x$

$=\int \cos ^{2} x\left[3-4\left(1-\cos ^{2} x\right)\right] \sin x d x$

$=\int \cos ^{2} x\left\{3-4+4 \cos ^{2} x\right\} \sin x d x$

$=\int \cos ^{2} x\left(4 \cos ^{2} x-1\right) \sin x d x$

Let z=cosx then dz=-sinxdx
-dz=sinxdx

Now , $\int \cos ^{2} x\left(4 \cos ^{2} x-1\right) \sin x d x$ $=\int z^{2}\left(4 z^{2}-1\right)(-d z)$

$=-\int\left(4 z^{4}-z^{2}\right) d z$

$=-4 \int z^{4} d 2+\int z^{2} d z$

$=-4 \frac{z^{5}}{5}+\frac{z^{3}}{3}+c$

$=-\frac{4}{5} \cos ^{5} x+\frac{\cos ^{3} x}{3}+c$

$=\frac{\cos ^{3} x}{3}-\frac{4}{5} \cos ^{5} x+c$

Question 14
$\int \cos ^{3} x \cdot e^{\log \sin x} d x$
Sol :
$=\int \cos ^{3} x \cdot \sin x d x$

$\left(\because e^{\log a}=a\right)$

Let z=cosx then dz=-sinxdx
-dz=sinxdx

Now , $\int \cos ^{3} x \cdot \sin x d x=\int z^{3}(-d z)=$ $=-\int z^{3}(d z)$

$=\frac{-z^{4}}{4}+c$

$=-\frac{\cos ^{4} x}{4}+c$

Question 15
$\int \sin ^{3} 2 x d x$
Sol :
$=\int \sin ^{2} 2 x \cdot \sin 2 x d x$

Let z=cos2x then dz=-2sin2xdx

Now , $\int \sin ^{2} 2 x \cdot \sin 2 x d x$

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

$=\int\left(1-z^{2}\right)\left(\frac{-d z}{2}\right)$

$=-\frac{1}{2} \int\left(1-z^{2}\right) d z$

$=-\frac{1}{2} \int d z+\frac{1}{2} \int z^{2} d z$

$=-\frac{1}{2} z+\frac{1}{2} \cdot \frac{z^{3}}{3}+c$

$=-\frac{\cos 2 x}{2}+\frac{1}{6} \cos ^{3} 2 x+c$
Question 16
$\int \cos ^{5} x \cdot \operatorname{cosec}^{2} x d x$
Sol :
$=\int \frac{\cos ^{2} x \cdot \cos ^{3} x}{\sin ^{2} x} d x$

$=\int \frac{\left(1-\sin ^{2} x\right)\left(1-\sin ^{2} x\right) \cdot \cos x}{\sin ^{2} x} d x$

Let z=sinx then dz=cosxdx

Now , $\int \frac{\left(1-z^{2}\right)\left(1-z^{2}\right) d z}{z^{2}}$

$=\int \frac{\left(1-z^{2}\right)^{2} d z}{z^{2}}$

$=\int \frac{1+z^{4}-2 z^{2}}{2^{z}} d z$

$=\int \frac{1}{z^{2}} d z+\int \frac{z^{4}}{z^{2}} d z-2 \int \frac{z^{2}}{z^{2}} d z$

$=\int z^{-2} d z+\int z^{2} d z-2 \int d z$

$=\frac{z^{-2+1}}{-2+1}+\frac{z^{2+1}}{2+1}-2 z+c$

$=-\frac{1}{z}+\frac{z^{3}}{3}-2z+c$

$=-\frac{1}{\sin x}+\frac{\sin ^{3} x}{3}-2 \sin x+c$

$=\frac{\sin ^{3} x}{3}-\operatorname{cosec} x-2 \sin x+c$

Question 17
$\int \cos ^{7} a x \cdot \sin a x d x$
Sol :
Let z=cosax then dz=-sinax.adx
$-\frac{d z}{a}=\sin a x d x$

Now , $\int \cos ^{7} a x \cdot \sin a x d x$ $=\int z^{7}\left(-\frac{d z}{a}\right)$
$=\frac{-1}{a} \int z^{7} d z$

$=-\frac{1}{a} \frac{z^{8}}{8}+c$

$=\frac{-z^{8}}{8 a}+c$

$=-\frac{\cos ^{8}(a x)}{8 a}+c$

Question 18
$\int \frac{d x}{\sin ^{2} x \cdot \cos ^{2} x}$
Sol :
$=\int \operatorname{cosec}^{2} x \cdot \sec ^{2} x d x$

$=\int\left(1+\cot ^{2} x\right) \cdot \sec ^{2} x d x$

$=\int\left(1+\frac{1}{\tan ^{2} x}\right) \cdot \sec ^{2} x d x$

$=\int\left(1+\cot ^{2} x\right) \cdot \sec ^{2} x d x$

$=\int\left(1+\frac{1}{\tan ^{2} x}\right) \cdot \sec ^{2} x d x$

$=\int\left(\frac{1+\tan ^{2} x}{\tan ^{2} x}\right) \cdot \sec ^{2} x d x$

Let z=tanx then $d z=\sec ^{2} x d x$

Now , $\int \frac{d x}{\sin ^{2} x \cdot \cos ^{2} x}=\int\left(\frac{1+\tan ^{2} x}{\tan ^{2} x}\right) \cdot \sec ^{2} x d x$

$=\int \frac{1+z^{2}}{z^{2}} \cdot d z$

$=\int \frac{1}{z^{2}}+\frac{z^{2}}{z^{2}} d z$

$=\int\left(z^{-2}+1\right) d z$

$=\int z^{-2} d 2+\int d z$

$=\frac{z^{-2 +1}}{-2+1}+z+c$

$=-\frac{1}{z}+z+c$

$=-\frac{1}{\tan x}+\tan x+c$

=-cotx+tanx+c

= tanx-cotx+c

Question 19
$\int \frac{d x}{\sin ^{3} x \cdot \cos ^{5} x}$
Sol :
$=\int \operatorname{cosec}^{3} x \cdot \sec ^{5} x d x$

$=\int \operatorname{cosec}^{2} x \cdot \operatorname{cosec} x \cdot \sec ^{2} x \cdot \sec ^{2} x \cdot \sec xdx$

$=\int \operatorname{cosec}^{2} x \cdot \sec ^{2} x \cdot \operatorname{cosec} x \cdot \sec x \cdot \sec ^{2} x d x$

$=\int\left(1+\cot ^{2} x\right)\left(1+\tan ^{2} x\right) \cdot \frac{2}{2 \sin x \cdot \cos x} \cdot \sec ^{2} x d x$

$=\int\left(1+\frac{1}{\tan ^{2} x}\right)\left(1+\tan ^{2} x\right) \cdot \frac{2}{\sin 2 x}.sec^2xdx$

$=\int\left(\frac{1+\tan ^{2} x}{\tan ^{2} x}\right)\left(1+\tan ^{2} x\right) \cdot \frac{2}{\left(\frac{2 \tan x}{1+\tan ^{2} x}\right)}.sec^2xdx$

$=\int\left(\frac{\left(1+\tan ^{2} x\right)}{\tan ^{2} x}\right)\left(1+\tan ^{2} x\right) \cdot 2\left(\frac{1+\tan ^{2} x}{2 \tan x}\right) \cdot \sec ^{2} x d x$

$=\int \frac{\left(1+\tan ^{2} x\right)^{3}}{\tan ^{3} x} \cdot \sec ^{2} x d x$

$\therefore \int \frac{d x}{\sin ^{3} x \cdot \cos ^{5} x}=\int \frac{\left(1+\tan ^{2} x\right)^{3}}{\tan ^{3} x} \cdot \sec ^{2} x d x$

Let z=tanx then $d z=\sec ^{2} x d x$

Now , $\int\left(\frac{1+z^{2}}{z^{2}}\right)^{3} d 2=\int \frac{\left(1+z^{6}+3z^{2}+3z^{4}\right)}{2^{3}} d z$

$=\int \frac{1}{z^{3}} d 2+\int \frac{z^{6}}{z^{3}} d 2+3 \int \frac{z^{2}}{z^{3}} d 2+3 \int \frac{z^{4}}{z^{3}} d 2$

$=\int z^{-3} d z+\int z^{3} d z+3 \int \frac{1}{z} d z+3 \int z d z$

$=\frac{z^{-3+1}}{-3+1}+\frac{z^{4}}{4}+3 \log |z|+\frac{3 z^{2}}{2} + c$

$=\frac{z^{-2}}{-2}+\frac{z^{4}}{4}+3 \log |2|+\frac{3}{2} \cdot z^{2}+c$

$=-\frac{1}{2 z^{2}}+\frac{z^{4}}{4}+3 \log |z|+\frac{3}{2} z^{2}+c$

$=-\frac{1}{2 \tan ^{2} x}+\frac{\tan ^{4} x}{4}+3 \log |\tan x|+\frac{3}{2} \tan ^{2} x+c$

$=-\frac{1}{2} \cot ^{2} x+3 \log |\tan x|+\frac{3}{2} \tan ^{2} x+\frac{\tan ^4x}{4}+c$

Question 20
$\int \frac{\sin ^{4} x}{\cos ^{6} x} d x$
Sol :
$=\int \frac{\sin ^{4} x}{\cos ^{4} x \cdot \cos ^{2} x} d x$

$=\int \tan ^{4} x \cdot \sec ^{2} x d x$

Let z=tanx then $d z=\sec ^{2} x d x$

Now , $\int \frac{\sin ^{4} x}{\cos ^{6} x} d x$

$=\int \tan ^{4} x \cdot \sec ^{2} x d x$

$=\int z^{4} d z$

$=\frac{z^{5}}{5}+c=\frac{\tan ^{5} x}{5}+c$

Question 21
$\int \sqrt{\frac{\cos x}{\sin ^{5} x}} d x$
Sol :
$=\int \sqrt{\frac{\cos x}{\sin ^{4} x \cdot \sin x}} d x$

$=\int \frac{1}{\sin ^{2} x} \cdot \sqrt{\frac{\cos x}{\sin x}} d x$

$=\int \operatorname{cosec}^{2} x \cdot \sqrt{\cot x} d x$

$=\int \sqrt{\cot x} \cdot \operatorname{cosec}^{2} x d x$

Let z=cotx then $d z=-\operatorname{cosec}^{2} x d x$
$-d z=\operatorname{cosec}^{2} x d x$

Now , $\int \sqrt{\frac{\cos x}{\sin ^{2} x}} d x=\int \sqrt{\cot x} \cdot \operatorname{cosec}^{2} x d x$

$=\int \sqrt{z}(-d z)=-\int \sqrt{z} d z$

$=-\int z^{\frac{1}{2}} d z$

$=-\frac{z^{\frac{1}{2}+1}}{\frac{-1}{2}+1}+c=\frac{-2}{3} z^{\frac{3}{2}}+c$

$=\frac{-2}{3} \cot ^{\frac{3}{2}} x+c$

Question 22
$\int \frac{d x}{4 \sin 2 x \cdot \cos ^{2} x}$
Sol :
$=\int \frac{d x \cdot \sec ^{2} x d x}{4\left(\frac{2 \tan x}{1+\tan ^{2} x}\right)}$

$=\frac{1}{8} \int \frac{\left(1+\tan ^{2} x\right)}{\tan x} \cdot \sec ^{2} x d x$

Let z=tanx then $d 2=\sec ^{2} x d x$

Now , $\int \frac{d x}{4 \sin 2 x \cdot \cos ^{2} x}$

$=\frac{1}{8} \int\left(\frac{1+tan^{2} x}{\tan x}\right) \cdot \sec ^{2} x d x$

$=\frac{1}{8} \int\left(\frac{1+z^{2}}{z}\right){d z}$

$=\frac{1}{8} \int \frac{1}{z} d z+\frac{1}{8} \int \frac{z^{2}}{z} d z$

$=\frac{1}{8} \log |2|+\frac{1}{8} \int z d z$

$=\frac{1}{8} \log | 2|+\frac{1}{8} \frac{z^{2}}{2}+c$

$=\frac{1}{8} \log |2|+\frac{1}{16} z^{2}+c$

$=\frac{1}{8} \log |\tan x|+\frac{1}{16} \tan ^{2} x+c$

Question 23

$\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x$

Sol :

$=\int \frac{2 \sqrt{\tan x}}{2 \sin x \cdot \cos x} d x$

$=\int \frac{2 \sqrt{\tan x}}{\sin 2 x} d x$

$=\int \frac{2 \sqrt{\tan x}}{\left(\frac{2 \tan x}{\left(1+\tan ^{2} x\right)}\right)} d x$

$=\int \frac{\left(1+\tan ^{2} x\right) \cdot \sqrt{\tan x}}{\tan x} d x$

$=\int \frac{\sqrt{\tan x} \cdot \sec ^{2} x d x}{\sqrt{\tan x} \cdot \sqrt{\tan x}}$

$=\int \frac{1 \cdot \sec ^{2} x d x}{\sqrt{\tan x}}$

Let z=tanx then $d z=\sec ^{2} x d x$

Now , $\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x$

$=\int \frac{\sec ^{2} x d x}{\sqrt{\tan x}}$

$=\int \frac{d z}{\sqrt{z}}$

$=\int \frac{1}{z^{\frac{1}{2}}} d z$

$=\int z^{-\frac{1}{2}} d z$

$=\frac{z^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}+c$

$=2 z^{\frac{1}{2}}+c$

$=2 \sqrt{z}+c$

$=2 \sqrt{\tan x}+c$

$\int \frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cdot \cos ^{2} x} d x$
Sol :
$=\int \frac{\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}}{\sin ^{2} x \cdot \cos ^{2} x} d x$

Now applying the formula a³+b³-=(a+b)³-3ab(a+b)

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

$=\int \frac{1-3 \sin ^{2} x \cdot \cos ^{2} x \cdot 1}{\sin ^{2} x \cdot \cos ^{2} x} d x$

$=\int \frac{1}{\sin ^{2} x \cdot \cos ^{2} x}-3 \int \frac{\sin ^{2} x \cdot \cos ^{2} x}{\sin ^{2} x \cos ^{2} x}$

$=\int \operatorname{cosec}^{2} x \cdot \sec ^{2} x d x-3 \int d x$

$\int\left(1+\cot ^{2} x\right) \cdot \sec ^{2} x d x-3 x+c$

$\int\left(1+\frac{1}{\tan ^{2} x}\right) \cdot \sec ^{2} x d x-3 x+c$

$\int\left(\frac{1+\tan ^{2} x}{\tan ^{2} x}\right) \cdot \sec ^{2} x d x-3 x+c$

Let z=tanx then $d z=\sec ^{2} x d x$

Now ,$\int\left(\frac{1+z^{2}}{z^{2}}\right) \cdot d z-3 x+c$

$=\int \frac{1}{z^{2}} d z+\int \frac{z^{2}}{z^{2}} d z-3 x+c$

$=\frac{z^{-2+1}}{-2+1}+z-3 x+c$

$=-z^{-1}+z-3 x+c$

$=-\tan ^{-1} x+\tan x-3 x+c$

$=-\frac{1}{\tan x}+\tan x-3 x+c$

=-cotx+tanx-3x+c

=tanx-cotx-3x+c

Question 25

$\int \cos ^{4} x d x$
Sol :
$=\int \cos ^{2} x \cdot \cos ^{2} x d x$

$=\int \frac{(1+\cos 2 x)}{2} \cdot \frac{(1+\cos 2 x)}{2} \cdot d x$

$\left(\because 1+\cos 2 x=2 \cos ^{2} x\right)$

$=\frac{1}{4} \int(1+\cos 2 x)^{2} d x$

$=\frac{1}{4} \int\left(1+\cos ^{2} 2 x+2 \cos 2 x\right) d x$

$=\frac{1}{4} \int d x+\frac{1}{4} \int \cos ^{2} 2 x d x+\frac{1}{4} \cdot 2 \int \cos 2 x d x$

$=\frac{1}{4} x+\frac{1}{4} \int \frac{(1+\cos 2 \cdot 2 x)}{2} d x+\frac{1}{2} \int \cos 2 x d x$

$=\frac{x}{4}+\frac{1}{8} \int d x+\frac{1}{8} \int \cos 4 x d x+\frac{1}{2} \int \cos 2 x d x$

$=\frac{x}{4}+\frac{x}{8}+\frac{1}{8} \frac{\sin 4 x}{4}+\frac{1}{2} \frac{\sin 2 x}{2}+c$

$=\frac{3 x}{8}+\frac{\sin 2 x}{4}+\frac{\sin 4 x}{32}+c$

Question 26

$\int \cos ^{4} 2 x d x$
Sol :
$=\int \cos ^{2} 2 x \cdot \cos ^{2} 2 x d x$

$=\int\left(\frac{1+\cos 2 \cdot 2 x}{2}\right)\left(\frac{1+\cos 2 \cdot 2 x}{2}\right) d x$

$=\frac{1}{4} \int(1+\cos 4 x)^{2} d x$

$=\frac{1}{4} \int\left(1+\cos ^{2} 4 x+2 \cos 4 x\right) d x$

$=\frac{1}{4} \int d x+\frac{1}{4} \int \cos ^{2} 4 x d x+\frac{1}{4} \cdot 2 \int \cos 4 x d x$

$=\frac{x}{4}+\frac{1}{4} \int\left(\frac{1+\cos 2 \cdot 4 x}{2}\right) d x+\frac{1}{2} \int \cos 4 x d x$

$=\frac{x}{4}+\frac{1}{8} \int d x+\frac{1}{8} \int \cos 8 x d x+\frac{1}{2} \int \cos 4 x d x+c$

$=\frac{x}{4}+\frac{x}{8}+\frac{1}{8} \frac{\sin 8 x}{8}+\frac{1}{2} \frac{\sin 4 x}{4}+c$

$=\frac{x}{4}+\frac{x}{8}+\frac{1}{8} \frac{\sin 8 x}{8}+\frac{1}{2} \frac{\sin 4 x}{4}+c$

$=\frac{3 x}{8}+\frac{1}{8} \frac{\sin 8 x}{8}+\frac{1}{8} \sin 4 x+c$

$=\frac{1}{8}\left[ 3 x+\sin 4 x+\frac{1}{8} \sin 8 x\right]+c$