# The value of the integral

Question:

The value of the integral $\int \frac{\sin \theta \cdot \sin 2 \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$ is

(where $c$ is a constant of integration)

1. (1) $\frac{1}{18}\left[9-2 \sin ^{6} \theta-3 \sin ^{4} \theta-6 \sin ^{2} \theta\right]^{\frac{3}{2}}+c$

2. (2) $\frac{1}{18}\left[11-18 \sin ^{2} \theta+9 \sin ^{4} \theta-2 \sin ^{6} \theta\right]^{\frac{3}{2}}+c$

3. (3) $\frac{1}{18}\left[11-18 \cos ^{2} \theta+9 \cos ^{4} \theta-2 \cos ^{6} \theta\right]^{\frac{3}{2}}+c$

4. (4) $\frac{1}{18}\left[9-2 \cos ^{6} \theta-3 \cos ^{4} \theta-6 \cos ^{2} \theta\right]^{\frac{3}{2}}+c$

Correct Option: , 3

Solution:

$\int \frac{2 \sin ^{2} \theta \cos \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{2 \sin ^{2} \theta} d \theta$

Let $\sin \theta=t, \cos \theta d \theta=d t$

$=\int\left(t^{6}+t^{4}+t^{2}\right) \sqrt{2 t^{4}+3 t^{2}+6} d t=\int\left(t^{5}+t^{3}+t\right) \sqrt{2 t^{6}+3 t^{4}+6 t^{2}} d t$

Let $2 t^{6}+3 t^{4}+6 t^{2}=z$

$12\left(t^{5}+t^{3}+t\right) d t=d z$

$=\frac{1}{12} \int \sqrt{z} d z=\frac{1}{18} z^{3 / 2}+c$

$=\frac{1}{18}\left[\left(2 \sin ^{6} \theta+3 \sin ^{4} \theta+6 \sin ^{2} \theta\right)^{3 / 2}+C\right.$

$=\frac{1}{18}\left[\left(1-\cos ^{2} \theta\right)\left(2\left(1-\cos ^{2} \theta\right)^{2}+3-3 \cos ^{2} \theta+6\right)\right]^{3 / 2}+C$

$=\frac{1}{18}\left[\left(1-\cos ^{2} \theta\right)\left(2 \cos ^{4} \theta-7 \cos ^{2} \theta+11\right)\right]^{3 / 2}+C$

$=\frac{1}{18}\left[-2 \cos ^{6} \theta+9 \cos ^{4} \theta-18 \cos ^{2} \theta+11\right]^{3 / 2}+C$