$\int\frac{\sin 2x\,dx}{\sqrt{9-\cos^4x}}$ equals

$-\sin^{-1}(\frac{\cos^2x}{3})+C$: C is an arbitrary constant

The correct answer is Option (4) → $-\sin^{-1}(\frac{\cos^2x}{3})+C$: C is an arbitrary constant

$I=\displaystyle\int \frac{\sin 2x\,dx}{\sqrt{9-\cos^{4}x}}$

Let $u=\cos^{2}x\;\Rightarrow\;du=-\sin 2x\,dx$.

$I=\displaystyle\int \frac{-\,du}{\sqrt{9-u^{2}}}=-\sin^{-1}\!\left(\frac{u}{3}\right)+C$

$\Rightarrow I=-\sin^{-1}\!\left(\frac{\cos^{2}x}{3}\right)+C$

$-\sin^{-1}\!\left(\frac{\cos^{2}x}{3}\right)+C$