int_{0}^{frac{pi}{2}} frac{sin x cos x}{1+sin | vedclass

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Question

(B) $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x$ $I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{2 \sin x \cos x}{1+(\sin ^{

Vedclass · Accepted Answer

$\frac{\pi}{8}$

Vedclass · Answer

(B) $I = \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x$ $I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{2 \sin x \cos x}{1+(\sin ^{2} x)^{2}} d x$ ધારો કે $\sin ^{2} x = t$,તેથી $2 \sin x \cos x d x = dt$. જ્યારે $x = 0$,$t = 0$ અને જ્યારે $x = \frac{\pi}{2}$,$t = 1$. $I = \frac{1}{2} \int_{0}^{1} \frac{dt}{1+t^{2}} = \frac{1}{2} [	an ^{-1} t]_{0}^{1}$ $I = \frac{1}{2} (	an ^{-1} 1 - 	an ^{-1} 0) = \frac{1}{2} (\frac{\pi}{4} - 0) = \frac{\pi}{8}$

$\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x=$

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