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$\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x$ is equal to
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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 $\sin ^{2} \mathrm{x}=\mathrm{t} \Rightarrow 2 \sin \mathrm{x} \cos \mathrm{xdx}=\mathrm{dt}$
$\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{2 \sqrt{1-\mathrm{t}^{2}}}$
$=\frac{1}{2} \sin ^{-1} \mathrm{t}+\mathrm{C}$
$=\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C$
Put $\sin ^{2} \mathrm{x}=\mathrm{t} \Rightarrow 2 \sin \mathrm{x} \cos \mathrm{xdx}=\mathrm{dt}$
$\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{2 \sqrt{1-\mathrm{t}^{2}}}$
$=\frac{1}{2} \sin ^{-1} \mathrm{t}+\mathrm{C}$
$=\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C$
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