int_0^(pi/2)sinxsin2xdx=? 

Explanation:

Another Explanation (5): সমাধান: \( \int_0^{\frac{\pi}{2}} \sin x \sin 2x \, dx \) আমরা জানি, \(\sin 2x = 2 \sin x \cos x\). সুতরাং, \(\int_0^{\frac{\pi}{2}} \sin x \sin 2x \, dx = \int_0^{\frac{\pi}{2}} \sin x (2 \sin x \cos x) \, dx \) \( = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \cos x \, dx \) এখন, ধরি \(u = \sin x\). তাহলে, \(du = \cos x \, dx\). যখন \(x = 0\), তখন \(u = \sin 0 = 0\). যখন \(x = \frac{\pi}{2}\), তখন \(u = \sin \frac{\pi}{2} = 1\). সুতরাং, ইন্টিগ্রালটি হবে: \(2 \int_0^1 u^2 \, du = 2 \left[ \frac{u^3}{3} \right]_0^1 = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 2 \left( \frac{1}{3} \right) = \frac{2}{3}\). অতএব, \( \int_0^{\frac{\pi}{2}} \sin x \sin 2x \, dx = \frac{2}{3} \). 🎉