int_0^{pi /2} frac{sin x cos x}{1 + sin^4 x} | vedclass

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(D) Let $I = \int_0^{\pi /2} \frac{\sin x \cos x}{1 + \sin^4 x} \, dx$. Substitute $\sin^2 x = t$. Then,$2 \sin x \cos x \, dx = dt$,which implies $\s

Vedclass · Accepted Answer

$\frac{\pi}{8}$

Vedclass · Answer

(D) Let $I = \int_0^{\pi /2} \frac{\sin x \cos x}{1 + \sin^4 x} \, dx$. Substitute $\sin^2 x = t$. Then,$2 \sin x \cos x \, dx = dt$,which implies $\sin x \cos x \, dx = \frac{1}{2} \, dt$. When $x = 0$,$t = \sin^2(0) = 0$. When $x = \frac{\pi}{2}$,$t = \sin^2(\frac{\pi}{2}) = 1$. Substituting these into the integral: $I = \int_0^1 \frac{1}{1 + t^2} \cdot \frac{1}{2} \, dt = \frac{1}{2} \int_0^1 \frac{1}{1 + t^2} \, dt$. $I = \frac{1}{2} [	an^{-1} t]_0^1$. $I = \frac{1}{2} (	an^{-1}(1) - 	an^{-1}(0)) = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}$.

$\int_0^{\pi /2} \frac{\sin x \cos x}{1 + \sin^4 x} \, dx = $

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