Evaluate the following integrals: x ^ -1 x 1-x^2 dx - Vidyadip

Question

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

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Answer

Let first function be $\sin^{-1}\text{x}$ and second dunction be $\frac{\text{x}}{\sqrt{1-\text{x}^2}}.$
First we find the intergral of the second function, i.e, $\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}.$
Put $\text{t}=1-\text{x}^{2}.$ Then $\text{dt}=-2\text{x dx}$
Therefore, $\int\frac{\text{x dx}}{\sqrt{1-\text{x}^2}}=-\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}}}=-\sqrt{\text{t}}=-\sqrt{1-\text{x}^2}$
Hence, $\int\frac{\text{x}\sin^{-1}}{\sqrt{1-\text{x}^2}}\text{dx}=(\sin^{-1}\text{x})\Big(-\sqrt{1-\text{x}^2}\Big)-\int\frac{1}{\sqrt{1-\text{x}^2}}\Big(-\sqrt{1-\text{x}^2}\Big)\text{dx}$
$=-\sqrt{1-\text{x}^2}\sin^{-1}\text{x}+\text{x + C}=\text{x}-\sqrt{1-\text{x}^2}\sin^{-1}\text{x+C}$

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