int_{-pi}^{frac{pi}{2}} sin x cdot sin^2(cos | vedclass

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Question

(C) Let $I = \int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx$. Substitute $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$

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$\frac{\sin 2-2}{4}$

Vedclass · Answer

(C) Let $I = \int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx$. Substitute $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$. When $x = -\pi$,$t = \cos(-\pi) = -1$. When $x = \frac{\pi}{2}$,$t = \cos(\frac{\pi}{2}) = 0$. Thus,$I = \int_{-1}^{0} \sin^2(t) (-dt) = \int_{0}^{-1} \sin^2(t) \, dt$. Using the identity $\sin^2(t) = \frac{1 - \cos(2t)}{2}$,we get: $I = \int_{0}^{-1} \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \left[ t - \frac{\sin(2t)}{2} \right]_{0}^{-1}$. $I = \frac{1}{2} \left[ (-1 - \frac{\sin(-2)}{2}) - (0 - 0) \right]$. Since $\sin(-2) = -\sin(2)$,we have: $I = \frac{1}{2} \left[ -1 + \frac{\sin(2)}{2} \right] = \frac{\sin(2) - 2}{4}$.

$\int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx =$

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