Question

Evaluate: $\int \frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{(1-x^2)}}dx$

• A. $\frac{1}{2}{\left({\sin }^{-1}x\right)}^4$
• B. ${\left({\sin }^{-1}x\right)}^4$
• C. $\frac{1}{3}{\left({\sin }^{-1}x\right)}^3$
• D. $\frac{1}{4}{\left({\sin }^{-1}x\right)}^4$

Answer 1

The correct option is D $\frac{1}{4}{\left({\sin }^{-1}x\right)}^4$

Let $I=\int \frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{(1-x^2)}}dx$

Put $t={\sin }^{-1}x\Rightarrow dt=\frac{1}{\sqrt{(1-x^2)}}dx$
Therefore
$I=\int t^{3}dt=\frac{t^4}{4}=\frac{{\left({\sin }^{-1}x\right)}^4}{4}$