Evaluate: sin^ -1 x dx. - Vidyadip

Question

Evaluate: $\int\text{x sin}^{-1}\text{x dx}.$

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Answer

$\int\text{x}\sin^{-1}\text{x dx}=\sin^{-1}\text{x}\cdot\frac{\text{x}^{2}}{2}-\int\frac{\text{x}^{2}}{2}\cdot\frac{1}{\sqrt{1 - \text{x}^{2}}}\text{dx}$
$=\frac{\text{x}^{2}}{2}\sin^{-1}\text{x}+\frac{1}{2}\int\frac{\text{1 - x}^{2}-1}{\sqrt{\text{1 - x}^{2}}}\text{dx}$
$=\frac{\text{x}^{2}}{2}\sin^{-1}\text{x}+\frac{1}{2}\int\sqrt{\text{1 - x}^{2}}\text{dx}-\frac{1}{2}\int\frac{\text{dx}}{\sqrt{\text{1 - x}^{2}}}$
$=\frac{\text{x}^{2}}{2}\sin^{-1}\text{x}+\frac{1}{2}\Bigg[\frac{\text{x}}{2}\sqrt{\text{1 - x}^{2}}+\frac{1}{2}\sin^{-1}\text{x}\Bigg]-\frac{1}{2}\sin^{-1}\text{x}+\text{c}$
$=\sin^{-1}\cdot\text{ x }\cdot\Bigg[\frac{\text{x}^{2}}{2}+\frac{1}{4}-\frac{1}{2}\Bigg]+\frac{1}{4}\text{ x }\sqrt{\text{1 - x}^{2}}+\text{c}$
$\text{ Or, }\frac{\text{2x}^{2}-1}{4}\sin^{-1}\text{x}+\frac{1}{4}\text{ x }\sqrt{\text{1 - x}^{2}}+\text{c}.$

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