Question

${\int }_0^1\frac{2x}{\sqrt{1-x^4}}dx$ is equal to?

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $2\pi$
• D. $0$

Answer 1

The correct option is B $\frac{\pi }{2}$

We have,

$I={\int }_0^1\frac{2x}{\sqrt{1-x^4}}dx$

Let

$t=x^2$

$dt=2xdx$

Therefore,

$I={\int }_0^1\frac{dt}{\sqrt{1-t^2}}dt$

$I={\left[{\sin }^{-1}t\right]}_0^1$

$I=[{\sin }^{-1}1-{\sin }^{-1}0]$

$I=[{\sin }^{-1}\sin \frac{\pi }{2}-{\sin }^{-1}\sin 0]$

$I=\frac{\pi }{2}$

Hence, this is the answer.