Question

$\int \frac{\sin 2x}{\sqrt{9-{\sin }^{4}x}}dx$ is equal to

• A. ${\tan }^{-1}\left(\frac{{\sin }^{2}x}{3}\right)+C$
• B. ${\sin }^{-1}\left(\frac{{\sin }^{2}x}{3}\right)+C$
• C. ${\tan }^{-1}\left(\frac{\sin x}{3}\right)+C$
• D. ${\sin }^{-1}\left(\frac{\sin x}{3}\right)+C$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{{\sin }^{2}x}{3}\right)+C$

We have,
$\int \frac{\sin 2x}{\sqrt{9-{\sin }^{4}x}}dx$
Let, ${\sin }^{2}x=t$
Differentiating both side, we get
$2\sin x\cos xdx=dt$
$\Rightarrow \sin 2xdx=dt$
$\Rightarrow \int \frac{dt}{\sqrt{9-t^2}}=\int \frac{dt}{\sqrt{3^2-t^2}}\{\because \int \frac{dt}{\sqrt{a^2-t^2}}={\sin }^{-1}\left(\frac{t}{a}\right)+C\}={\sin }^{-1}\left(\frac{t}{3}\right)+C={\sin }^{-1}\left(\frac{{\sin }^{2}x}{3}\right)+C$