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$\int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin ^2(\cos x) d x=$
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Verified Answer
The correct answer is:
$\frac{\sin 2-2}{4}$
$I=\int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin ^2(\cos x) d x$
Let $\cos x=t \Rightarrow-\sin x d x=d t$
$\begin{aligned}
& \therefore \mathrm{I}=\int_{-1}^0 \sin ^2 t(-d t)=-\int_{-1}^0 \frac{1}{2}(1-\cos 2 x) d x \\
& =-\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{-1}^0=-\frac{1}{2}\left[0-(-1)+\frac{\sin (-2)}{2}\right] \\
& =\frac{1}{2}\left[\frac{\sin (2)}{2}-1\right] \\
& =-\frac{1}{2}\left[\frac{\sin (2)-2}{2}\right]=\frac{\sin 2-2}{4}
\end{aligned}$
Let $\cos x=t \Rightarrow-\sin x d x=d t$
$\begin{aligned}
& \therefore \mathrm{I}=\int_{-1}^0 \sin ^2 t(-d t)=-\int_{-1}^0 \frac{1}{2}(1-\cos 2 x) d x \\
& =-\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_{-1}^0=-\frac{1}{2}\left[0-(-1)+\frac{\sin (-2)}{2}\right] \\
& =\frac{1}{2}\left[\frac{\sin (2)}{2}-1\right] \\
& =-\frac{1}{2}\left[\frac{\sin (2)-2}{2}\right]=\frac{\sin 2-2}{4}
\end{aligned}$
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