The correct evaluation of int_0^{pi /2} {sin | vedclass

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Question

(D) Let $I = \int_0^{\pi /2} {\sin x \sin 2x \, dx}$.Using the trigonometric identity $\sin 2x = 2 \sin x \cos x$,we get:$I = \int_0^{\pi /2} {\sin x

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$\frac{2}{3}$

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(D) Let $I = \int_0^{\pi /2} {\sin x \sin 2x \, dx}$.Using the trigonometric identity $\sin 2x = 2 \sin x \cos x$,we get:$I = \int_0^{\pi /2} {\sin x (2 \sin x \cos x) \, dx} = 2 \int_0^{\pi /2} {\sin^2 x \cos x \, dx}$.Let $t = \sin x$. Then $dt = \cos x \, dx$.When $x = 0$,$t = \sin 0 = 0$. When $x = \pi / 2$,$t = \sin(\pi / 2) = 1$.Substituting these into the integral:$I = 2 \int_0^1 {t^2 \, dt} = 2 \left[ \frac{t^3}{3} \right]_0^1 = 2 \left( \frac{1}{3} - 0 \right) = \frac{2}{3}$.

The correct evaluation of $\int_0^{\pi /2} {\sin x\,\sin 2x} \, dx$ is

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