The value of the integral int frac{sin theta | vedclass

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(D) Let $I = \int \frac{\sin 	heta \cdot \sin 2 	heta \left(\sin ^{6} 	heta+\sin ^{4} 	heta+\sin ^{2} 	hetaight) \sqrt{2 \sin ^{4} 	heta+3 \si

Vedclass · Accepted Answer

$\frac{1}{18}\left[11-18 \cos ^{2} 	heta+9 \cos ^{4} 	heta-2 \cos ^{6} 	hetaight]^{\frac{3}{2}}+ c$

Vedclass · Answer

(D) Let $I = \int \frac{\sin 	heta \cdot \sin 2 	heta \left(\sin ^{6} 	heta+\sin ^{4} 	heta+\sin ^{2} 	hetaight) \sqrt{2 \sin ^{4} 	heta+3 \sin ^{2} 	heta+6}}{1-\cos 2 	heta} d 	heta$.Using $\sin 2 	heta = 2 \sin 	heta \cos 	heta$ and $1 - \cos 2 	heta = 2 \sin ^{2} 	heta$,we get:$I = \int \frac{\sin 	heta \cdot (2 \sin 	heta \cos 	heta) \cdot \sin ^{2} 	heta (\sin ^{4} 	heta + \sin ^{2} 	heta + 1) \sqrt{2 \sin ^{4} 	heta + 3 \sin ^{2} 	heta + 6}}{2 \sin ^{2} 	heta} d 	heta$.$I = \int \sin ^{2} 	heta \cos 	heta (\sin ^{4} 	heta + \sin ^{2} 	heta + 1) \sqrt{2 \sin ^{4} 	heta + 3 \sin ^{2} 	heta + 6} d 	heta$.Let $t = \sin 	heta$,then $dt = \cos 	heta d 	heta$.$I = \int t^{2} (t^{4} + t^{2} + 1) \sqrt{2 t^{4} + 3 t^{2} + 6} dt = \int (t^{6} + t^{4} + t^{2}) \sqrt{2 t^{4} + 3 t^{2} + 6} dt$.This simplifies to $\int (t^{5} + t^{3} + t) \sqrt{2 t^{6} + 3 t^{4} + 6 t^{2}} dt$.Let $u^{2} = 2 t^{6} + 3 t^{4} + 6 t^{2}$,then $2u du = (12 t^{5} + 12 t^{3} + 12 t) dt = 12(t^{5} + t^{3} + t) dt$.So,$(t^{5} + t^{3} + t) dt = \frac{u du}{6}$.$I = \int u \cdot \frac{u du}{6} = \frac{1}{6} \int u^{2} du = \frac{u^{3}}{18} + c = \frac{(2 t^{6} + 3 t^{4} + 6 t^{2})^{3/2}}{18} + c$.Substituting $t^{2} = 1 - \cos^{2} 	heta$,we get $2(1-\cos^{2}	heta)^{3} + 3(1-\cos^{2}	heta)^{2} + 6(1-\cos^{2}	heta)$.Expanding this leads to $11 - 18 \cos^{2} 	heta + 9 \cos^{4} 	heta - 2 \cos^{6} 	heta$. Thus,option $D$ is correct.

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)

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