Question
Evaluate the following integrals:
$\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$

Answer

$\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$

Let $\sin^{-1}\text{x}=\text{t}$

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

Now, $\int\frac{(\sin^{-1}\text{x})^3}{\sqrt{1-\text{x}^2}}=\text{dx}$

$=\int\text{t}^3\text{ dt}$

$=\frac{\text{t}^4}{4}+\text{C}$

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

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