∫0π2(1+cos⁡x)2sin⁡x dx=? \int_{0}^{\frac{\pi}{2}} (1 + \cos x)^2 \sin x \, dx = ?

Question

∫0π2(1+cos⁡x)2sin⁡x dx=? \int_{0}^{\frac{\pi}{2}} (1 + \cos x)^2 \sin x \, dx = ?

Accepted Answer

ক. 73 \frac{7}{3}

ব্যাখ্যা: ধরি, $u = 1 + \cos x$। অতএব, $du = -\sin x \, dx \implies \sin x \, dx = -du$।
যখন $x = 0$, $u = 1 + 1 = 2$; যখন $x = \frac{\pi}{2}$, $u = 1 + 0 = 1$।
$\int_{0}^{\frac{\pi}{2}} (1 + \cos x)^2 \sin x \, dx = \int_{2}^{1} u^2 (-du) = \int_{1}^{2} u^2 \, du = \left[ \frac{u^3}{3} \right]_1^2 = \frac{8 - 1}{3} = \frac{7}{3}$।

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