$ \int_0^x \frac{x}{\sin x}\left(3 \cos ^2 x+2 \sin x+\sin ^3 x-3\right) d x= $

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { given } \int_0^\pi \frac{x}{\sin x}\left(3 \cos ^2 x+2 \sin x+\sin ^3 x-3ight) d x \ & \Rightarrow \int_0^\pi \frac{x}{\sin x}\left(-3 \sin ^2 x+2 \sin x+\sin ^3 xight) d x \ & \Rightarrow \int_0^\pi x\left(\sin ^2 x-3 \sin x+2ight) d x\end{aligned}$ Let $I=\int_0^\pi x\left(\sin ^2 x-3 \sin x+2ight) d x ...(1)$ using propert $(a+b-x)$ $ I=\int_0^\pi(\pi-x)\left(\sin ^2 x-3 \sin x+2ight) d x ...(ii) $ adding (1) and (ii) $ \begin{aligned} & 2 I=\int_0^\pi \pi\left(\sin ^2(x)-3 \sin x+2ight) d x \ & 2 I=\int_0^\pi \pi\left(\frac{1-\cos 2 x}{2}-3 \sin x+2ight) d x \ & 2 I=\frac{\pi}{2} \int_0^\pi\left(5-\cos ^2 x-6 \sin xight) d x \ & I=\frac{\pi}{4}\left[5 x-\frac{\sin 2 x}{2}+6 \cos ^xight]_0^\pi \ & =\frac{\pi}{4}\left[\left(5 x \frac{-1}{2} \sin 2 \pi+6 \cos \piight)-(0-0+6)ight] \ & =\frac{\pi}{4}[5 \pi 6-6] \ & =\frac{\pi}{4}[5 \pi-12] \end{aligned} $

$ \int_0^x \frac{x}{\sin x}\left(3 \cos ^2 x+2 \sin x+\sin ^3 x-3\right) d x= $

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