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Question:

$x \sin ^{-1} x$

Solution:

Let $I=\int x \sin ^{-1} x d x$

Taking $\sin ^{-1} x$ as first function and $x$ as second function and integrating by parts, we obtain

$I=\sin ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \sin ^{-1} x\right) \int x d x\right\} d x$

$=\sin ^{-1} x\left(\frac{x^{2}}{2}\right)-\int \frac{1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x$

$=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x$

$=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\frac{1-x^{2}}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}\right\} d x$

$=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\sqrt{1-x^{2}}-\frac{1}{\sqrt{1-x^{2}}}\right\} d x$

$=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\left\{\int \sqrt{1-x^{2}} d x-\int \frac{1}{\sqrt{1-x^{2}}} d x\right\}$

$=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\left\{\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\sin ^{-1} x\right\}+\mathrm{C}$

$=\frac{x^{2} \sin ^{-1} x}{2}+\frac{x}{4} \sqrt{1-x^{2}}+\frac{1}{4} \sin ^{-1} x-\frac{1}{2} \sin ^{-1} x+\mathrm{C}$

$=\frac{1}{4}\left(2 x^{2}-1\right) \sin ^{-1} x+\frac{x}{4} \sqrt{1-x^{2}}+\mathrm{C}$