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$\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x=$
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Verified Answer
The correct answer is:
$\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C$
Let $I=\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x$
Put $\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t$
$\begin{aligned} \therefore \quad I &=\frac{1}{2} \int \frac{d t}{\sqrt{1-t^{2}}} \\ &=\frac{1}{2} \sin ^{-1} t+C=\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C \end{aligned}$
Put $\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t$
$\begin{aligned} \therefore \quad I &=\frac{1}{2} \int \frac{d t}{\sqrt{1-t^{2}}} \\ &=\frac{1}{2} \sin ^{-1} t+C=\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C \end{aligned}$
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