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  4. The value of the integral ∫ ( sin θ ⋅ sin 2 θ( sin 6 θ+ sin 4 θ+ sin 2 θ) √2 sin 4 θ+3 sin 2 θ+6/1- cos 2 θ) d θ is: (where c is a constant of integration)

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Q. 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)

JEE MainJEE Main 2021Integrals

Solution:

$I =\int \frac{\sin \theta . \sin 2 \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$
$\Rightarrow I =\int \frac{\sin \theta .2 \sin \theta \cos \theta \cdot \sin ^{2} \theta\left(\sin ^{4} \theta+\sin ^{2} \theta+1\right)\left(2 \sin ^{4} \theta+3 \sin ^{2} \theta+6\right)^{1 / 2}}{2 \sin ^{2} \theta} d \theta$
$=\int \sin ^{2} \theta \cdot \cos \theta\left(\sin ^{4} \theta+\sin ^{2} \theta+1\right)\left(2 \sin ^{4} \theta+3 \sin ^{2} \theta+6\right)^{1 / 2} d \theta$
Let $ \sin \theta= t $
$\Rightarrow \cos \theta d \theta= dt $
$\therefore I =\int t ^{2}\left( t ^{4}+ t ^{2}+1\right)\left(2 t ^{4}+3 t ^{2}+6\right)^{1 / 2} dt $
$=\int\left( t ^{5}+ t ^{3}+ t \right) t \left(2 t ^{4}+3 t ^{2}+6\right)^{1 / 2} dt$
$=\int\left( t ^{5}+ t ^{3}+ t \right)\left( t ^{2}\right)^{1 / 2}\left(2 t ^{4}+3 t ^{2}+6\right)^{1 / 2} dt$
$=\int\left( t ^{5}+ t ^{3}+ t \right)\left(2 t ^{6}+3 t ^{4}+6 t ^{2}\right)^{1 / 2} dt$
Let $2 t^{6}+3 t^{4}+6 t^{2}=u^{2} $
$\Rightarrow 12\left(t^{5}+t^{3}+t\right) d t=2udu $
$\therefore I=\int\left(u^{2}\right)^{1 / 2} \cdot \frac{2udu}{12} $
$=\int \frac{u^{2}}{6} d u=\frac{u^{3}}{18}+C$
$=\frac{\left(2 t^{6}+3 t^{4}+6 t^{2}\right)^{3 / 2}}{18}+C $
when $ t=sin \theta $
and $t^{2}=1-cos ^{2} \theta$