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KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals Exercise 19.4

Exercise 19.4

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


Question 24
\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

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