Question

$\int \frac{\sin x\cos x}{\sqrt{1-{\sin }^{4}x}}dx=$

• A. $\frac{1}{2}{\sin }^{-1}\left({\sin }^{2}x\right)+C$
• B. $\frac{1}{2}{\cos }^{-1}\left({\sin }^{2}x\right)+C$
• C. ${\tan }^{-1}\left({\sin }^{2}x\right)+C$
• D. ${\tan }^{-1}\left(2\sin x\right)+C$

Answer 1

The correct option is A $\frac{1}{2}{\sin }^{-1}\left({\sin }^{2}x\right)+C$

Let $I=\int \frac{\sin x.\cos x.dx}{\sqrt{1-{\sin }^{4}x}}$

Let, $y={\sin }^{2}x⟹dy=2\sin x.\cos x.dx$

$\therefore I=\frac{1}{2}\int \frac{1}{\sqrt{1-y^2}}dy$

$=\frac{{\sin }^{-1}y}{2}+C$

$=\frac{{\sin }^{-1}\left({\sin }^{2}x\right)}{2}+C$