Question

Evaluate : $\int \frac{x{\sin }^{-1}\left(x^2\right)}{\sqrt{1-x^4}}dx$

• A. $\frac{1}{4}{\left({\sin }^{-1}{x}^2\right)}^2+c$
• B. $\frac{1}{3}{\left({\sin }^{-1}{x}^2\right)}^2+c$
• C. $\frac{1}{2}{\left({\sin }^{-1}{x}^2\right)}^2+c$
• D. None

Answer 1

Answer: (A)

$\frac{1}{4}{\left({\sin }^{-1}{x}^2\right)}^2+c$

We have,

$I=\int \frac{x{\sin }^{-1}\left(x^2\right)}{\sqrt{1-x^4}}dx$

Let

$t={\sin }^{-1}{x}^2$

$\frac{dt}{dx}=\frac{1}{\sqrt{1-(x^2{)}^2}}\times \left(2x\right)$

$\frac{dt}{2}=\frac{x}{\sqrt{1-x^4}}dx$

Therefore,

$I=\frac{1}{2}\int tdt$

$I=\frac{1}{2}\frac{t^2}{2}+C$

$I=\frac{t^2}{4}+C$

On putting the value of $t$, we get

$I=\frac{({\sin }^{-1}x{)}^2}{4}+C$

Hence, this is the answer.