Question

$\int \frac{Sin \, x \, Cos \, x}{\sqrt{1 - Sin^4x}} dx$ =

• A. $\frac{1}{2} \; Sin^{-1} (Sin^2x)+C$
• B. $\frac{1}{2} Cos^{-1} (Sin^2x)+C$
• C. $Tan^{-1} (Sin^2x)+C$
• D. $Tan^{-1} (2 \, Sin \, x) + C$

Answer 1

Answer: (A)

Let $\,\,\,\,\,I=\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x$
Put $\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t$
$ \therefore \,\,\,\,\,I=\int \frac{d t}{2 \sqrt{1-t^{2}}}$
$=\frac{1}{2} \sin ^{-1} t+C$
$=\frac{1}{2} \sin ^{-1}\left(\sin ^{2} x\right)+C$