Question

$\int \frac{\sin x\cos x}{\sqrt{1-{\sin }^{4}x}}dx$ is equal to

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

Answer 1

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

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

Put $\sin x=t\Rightarrow \cos xdx=dt$

$\Rightarrow \int \frac{tdt}{\sqrt{1-t^4}}$

Again, put $t^2=y\Rightarrow 2tdt=dy\Rightarrow tdt=\frac{dy}{2}$

$\Rightarrow \int \frac{1}{\sqrt{1-y^2}}\frac{dy}{2}$

$=\frac{1}{2}\int \frac{1}{\sqrt{1-y^2}}dy$ Again, put $y=\sin \theta$

$\Rightarrow dy=\cos \theta d\theta$

$=\frac{1}{2}\int \frac{1\left(\cos \theta \right)}{\sqrt{1-{\sin }^2\theta }}d\theta$

$=\frac{1}{2}\int \frac{\cos \theta }{\cos \theta }d\theta =\frac{1}{2}\int d\theta =\frac{1}{2}\theta$

Putting back the substituted values, we get

$=\frac{1}{2}{\sin }^{-1}\left(y\right)$

$=\frac{1}{2}{\sin }^{-1}\left(t^2\right)$

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