Explanation: 
Another Explanation (5):
সমাধান:
আমরা জানি, \( \sin 2x = 2 \sin x \cos x \).
অতএব,
\[
\int_{0}^{\pi/2} \sin x \sin 2x \, dx = \int_{0}^{\pi/2} \sin x (2 \sin x \cos x) \, dx
\]
\[
= 2 \int_{0}^{\pi/2} \sin^2 x \cos x \, dx
\]
এখন, ধরি \( u = \sin x \), তাহলে \( du = \cos x \, dx \).
যখন \( x = 0 \), \( u = \sin 0 = 0 \).
যখন \( x = \pi/2 \), \( u = \sin (\pi/2) = 1 \).
সুতরাং,
\[
2 \int_{0}^{\pi/2} \sin^2 x \cos x \, dx = 2 \int_{0}^{1} u^2 \, du
\]
\[
= 2 \left[ \frac{u^3}{3} \right]_{0}^{1} = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right)
\]
\[
= 2 \left( \frac{1}{3} - 0 \right) = \frac{2}{3}
\]
অতএব, \( \int_{0}^{\pi/2} \sin x \sin 2x \, dx = \frac{2}{3} \). 🎉
