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KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals Exercise 19.5

Exercise 19.5

Question 1 
\int \tan x \cdot \sec ^{2} x d x
Sol :
Let z=tanx then d z=\sec ^{2} x d x

Now , \int \tan x \cdot \sec ^{2} x d x

=\int z d z=\frac{z^{2}}{2}+c

=\frac{\tan ^{2} x}{2}+c


Question 2
\int \tan ^{5} \theta \cdot \sec ^{4} \theta d \theta
Sol :
Let z=tanθ then d z=\sec ^{2} \theta d \theta

Now , \int \tan ^{5} \theta \cdot \sec ^{4} \theta d \theta

=\int \tan ^{5} \theta \cdot \sec ^{2} \theta \cdot \sec ^{2} \theta d \theta

=\int \tan ^{5} \theta \cdot\left(1+\tan ^{2} \theta\right) \cdot \sec ^{2} \theta d \theta

=\int z^{5}\left(1+z^{2}\right) d z

=\int\left(z^{5}+z^{7}\right) d z

=\frac{z^{6}}{6}+\frac{z^{8}}{8}+c

=\frac{\tan ^{6} \theta}{6}+\frac{\tan ^{8} \theta}{8}+c


Question 3
\int \tan ^{4} x \cdot \sec ^{2} x d x
Sol :
Let z=tanx d z=\sec ^{2} x d x

Now , \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 4
\int \sec ^{6} x d x
Sol :
Let z=tanx then d z=\sec ^{2} x d x

Now , \int \sec ^{6} x d x

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

=\int \sec ^{2} x \cdot \sec ^{2} x \cdot \sec ^{2} x d x

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

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

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

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

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

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

=2+\frac{z^{5}}{5}+2 \frac{z^{3}}{3}+c

=\tan x+\frac{\tan ^{5} x}{5}+\frac{2}{3} \tan ^{3} x+c

\tan x+\frac{2}{3} \tan ^{3} x+\frac{\tan ^{5} x}{5}+c


Question 5
\int \sec ^{4} \frac{x}{2} d x
Sol :
Let z=\tan \frac{x}{2} then d z=\sec ^{2} \frac{x}{2} \times \frac{1}{2} d x

2 d z=\sec ^{2} \frac{x}{2} d x

Now , \int \sec ^{4} \frac{x}{2} d x

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

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

=\int\left(1+z^{2}\right)(zd z)

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

=2\left[2+\frac{z^{3}}{3}\right]+c

=2 z+\frac{2}{3} z^{3}+c

=2 \tan \frac{x}{2}+\frac{2}{3} \tan^3 \frac{x}{2}+c

=2 \tan \frac{x}{2}\left[1+\frac{1}{3} \tan ^{2} \frac{x}{2}\right]+c


Question 6
\int \sec ^{3} x \cdot \tan x d x
Sol :
Let z=secx then  dz=secx.tanxdx

Now , \int \sec ^{3} x \cdot \tan x d x 

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

=\int z^{2} d z

=\frac{z^{3}}{3}+c=\frac{\sec ^{3} x}{3}+c


Question 7
\int \sec ^{6} x \cdot \tan x d x
Sol :
Let z=tanx then d z=\sec ^{2} x d x

Now , \int \sec ^{6} x \cdot \tan x d x

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

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

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

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

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

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

=\int\left(z+z^{5}+2 z^{3}\right) d z

=\int z d z+\int z^{5} d z+2 \int z^{3} d z

=\frac{z^{2}}{2}+\frac{z^{6}}{6}+\frac{1}{2} z^{4}+c

=\frac{\tan ^{2} x}{2}+\frac{\tan ^{6} x}{6}+\frac{1}{2} \tan ^{4} x+c

Question 8
(i) \int \sec ^{n} x \cdot \tan x d x
Sol :
Let z=secx then dz=secx.tanx.dx

Now , \int \sec ^{n} x \cdot \tan x d x

=\int \sec ^{n-1} \sec ^{1} x \cdot \tan x d x

=\int \sec ^{n-1} x \cdot \sec x \cdot \tan x d x

=\int z^{n-1} d z

=\frac{z^{n-1+1}}{n-1+1}+c

=\frac{z^{n}}{n}+c

=\frac{\sec ^{n} x}{n}+c


(ii) \int \sec ^{3} x \tan x d x
Sol :
Let z=secx then dz=secx.tanxdx

Now , \int \sec ^{3} x \cdot \tan x d x

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

=\int z^{2} d z=\frac{z^{3}}{3}+c

=\frac{\sec ^{3} x}{3}+c


(iii) \int \tan ^{3} 2 x \cdot \sec 2 x d x
Sol :
Let z=sec2x then dz=2sec2x.tan2xdx

\frac{d z}{2}=\sec 2 x \cdot \tan 2 x d x

Now , \int \tan ^{3} 2 x \cdot \sec 2 x d x

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

 \int\left(\sec ^{2} 2 x-1\right) \cdot \sec 2 x \cdot \tan 2 x d x

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

=\frac{1}{2}\left[\frac{z^{3}}{3}-z\right]+c

=\frac{1}{2}\left[\frac{\sec ^{3} 2 x}{3}-\sec 2 x\right]+c

Question 9
\int \sec ^{\frac{7}{2}} x \cdot \tan x d x
Sol :
Let z=secx then dz=secx.tanx.dx

Now , \int \sec ^{\frac{7}{2}} x \cdot \tan x d x

=\int \sec ^{\frac{5}{2}} x \cdot \sec ^{1} x \cdot \tan x d x

=\int z^{\frac{5}{2}} d z=\frac{z^{\frac{5}{2}+1}}{\frac{5}{2}+1}+c

=\frac{2^{\frac{7}{2}}}{\frac{7}{2}}+c

=\frac{2}{7} z^{\frac{7}{2}}+c

=\frac{2}{7} \cdot \sec ^{\frac{7}{2}} x+c


Question 10
\int \sec^4 \frac{x}{2} \cdot \tan \frac{x}{2} d x
Sol :
Let z=\sec \frac{x}{2} then

d z=\frac{1}{2} \sec \frac{x}{2} \cdot \tan \frac{x}{2} d x

2 d z=\sec \frac{x}{2} \cdot \tan \frac{x}{2} d x

Now , \int \sec^4 \frac{x}{2} \cdot \tan \frac{x}{2} d x

=\int \sec ^{3} \frac{x}{2} \cdot \sec \frac{x}{2} \cdot \tan \frac{x}{2} d x

=\int z^32 d z=

=2 \int z^{3} d z

=2 \cdot \frac{z^{4}}{4}+c

=\frac{1}{2} z^{4}+c

=\frac{1}{2} \cdot \sec^4 \frac{x}{2}+c

Question 11
\int \tan ^{5} x d x
Sol :
Let z=secx then dz=sec.tanx.dx

Now , \int \tan ^{4} x \cdot \tan x d x=\int \frac{\left(\tan ^{2} x\right)^{2} \cdot \sec x \cdot \tan x}{\sec x} d x

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

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

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

=\int\left(z^{3}+\frac{1}{z}-2 z\right) d z

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

=\frac{\sec ^{4} x}{4}+\log |\sec x|-\sec ^{2} x+c

=\frac{\sec ^{4} x}{4}-\sec ^{2} x+\log |\sec x|+c


Question 12
\int \cot ^{n} x \cdot \operatorname{cosec}^{2} x d x, \quad n \neq-1
Sol :
Let z=cotx d z=-\operatorname{cosec}^{2} x d x
-d z=\operatorname{cosec}^{2} x d x

Now , \int \cot ^{n} x \cdot \operatorname{cosec}^{2} x d x

=\int z^{n}(-d z)=-\int z^{n} d z

=-\frac{z^{n+1}}{n+1}+c

=-\frac{c o t^{n+1}}{n+1}+c


Question 13
\int \sqrt{\cot x} \cdot \operatorname{cosec}^{2} x d x
Sol :
Let z=cotx then d z=-\operatorname{cosec}^{2} x d x
-d z=\operatorname{cosec}^{2} x d x

Now , \int \cot ^{n} x \cdot \operatorname{cosec}^{2} x d x

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

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

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

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

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

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

Question 14
\int \cot x d x
Sol :
=\int \frac{\cos x d x}{\sin x}

Let z=sinx then dz=cosxdx

Now , \int \cot x d x=\int \frac{\cos x d x}{\sin x} 

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

=log|z|+c

=log|sinx|+c

=log|sinx|+c

=\log \left|\frac{1}{\operatorname{cosec} x}\right|+c

=log1-log|cosecx|+c

=0-log|cosecx|+c

[log|1|=0]

=-log|cosecx|+c


Question 15
\int \cot ^{3} x d x
Sol :
 Let z=cosecx then dz=-cosecx.cotx.dx
-dz=cosecx.cotxdx

Now , \int \cot ^{3} x d x

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

=\int \frac{\left(\operatorname{cosec}^{2} x-1\right) \cdot \operatorname{cosec} x \cdot \cot x d x}{\operatorname{cosec} x}

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

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

=-\frac{z^{2}}{2}+\log |z|+c

=-\frac{\operatorname{cosec}^{2} x}{2}+\log |\operatorname{cosec} x|+c

Question 16
\int \cot ^{4} x d x
Sol :
=\int\left(\cot ^{2} x\right)^{2} d x=\int\left(\operatorname{cosec}^{2} x-1\right)^{2} d x

=\int\left(\operatorname{cosec}^{4} x+1-2 \operatorname{cosec}^{2} x\right) d x

=\int \operatorname{cosec}^{4} x d x+\int d x-2 \int \operatorname{cosec}^{2} x d x

=\int \operatorname{cosec}^{2} x \cdot \operatorname{cosec}^{2} x d x+x-2(-\cot x)+c

=\int\left(1+\cot ^{2} x\right) \cdot \operatorname{cosec}^{2} x d x+x+2 \cot x+c

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

Now , \int\left(1+\cot ^{2} x\right) \cdot \operatorname{cosec}^{2} x d x+x+2cot x+c

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

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

=-\left[z+\frac{z^{3}}{3}\right]+x+2 \cot x+c

-\left[\cot x+\frac{\cot ^{3} x}{3}\right]+x+2 \cot x+c

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

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

Question 17
\int \frac{\sin ^{2} x}{(1+\cos x)^{2}} d x
Sol :
=\int \frac{1-\cos ^{2} x}{(1+\cos x)^{2}} d x

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

=\int \frac{1-\cos x}{1+\cos x} d x

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

=\int \tan^2 \frac{ x}{2} d x

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

=2 \tan \frac{x}{2}-x+c

=2 \tan \frac{x}{2}-x+c



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