Question

$\int \frac{SinxCosx}{\sqrt{1-Si{n}^{4}x}}dx$ =

• A. $\frac{1}{2}Si{n}^{-1}(Si{n}^{2}x)+C$
• B. $\frac{1}{2}Co{s}^{-1}(Si{n}^{2}x)+C$
• C. $Ta{n}^{-1}(Si{n}^{2}x)+C$
• D. $Ta{n}^{-1}(2Sinx)+C$

Answer 1

Answer: (A)

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