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$\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x=$
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Verified Answer
The correct answer is:
$\frac{\pi}{8}$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x \quad=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{2 \sin x \cos x}{1+\left(\sin ^{2} x\right)^{2}} d x$
Put $\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t$
When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=1$
$\therefore I=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}=\frac{1}{2}\left[\tan ^{-1} \mathrm{t}\right]_{0}^{1}=\frac{1}{2}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{8}$
Put $\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t$
When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=1$
$\therefore I=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}=\frac{1}{2}\left[\tan ^{-1} \mathrm{t}\right]_{0}^{1}=\frac{1}{2}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{8}$
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