∫xsin−1⁡xdx is equal to

# $\int {\mathrm{xsin}}^{-1}\mathrm{xdx}$ is equal to

1. A

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

2. B

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

3. C

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

4. D

None of the above

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### Solution:

Let
On taking sin-1 x as first function and x as second function and integrating by parts, we get
[$\because$ inverse functions comes before,algebraic functions in ILATE]

$=\frac{{\mathrm{x}}^{2}}{2}\cdot {\mathrm{sin}}^{-1}\mathrm{x}-\int \left[\frac{1-\left(1-{\mathrm{x}}^{2}\right)}{\sqrt{1-{\mathrm{x}}^{2}}}\cdot \frac{1}{2}\right]\mathrm{dx}$
[add and subtract 1 in numerator of second term]
$\begin{array}{l}=\frac{{\mathrm{x}}^{2}}{2}\cdot {\mathrm{sin}}^{-1}\mathrm{x}-\frac{1}{2}\int \frac{1}{\sqrt{1-{\mathrm{x}}^{2}}}\mathrm{dx}+\frac{1}{2}\int \frac{1-{\mathrm{x}}^{2}}{\sqrt{1-{\mathrm{x}}^{2}}}\mathrm{dx}\\ =\frac{{\mathrm{x}}^{2}}{2}\cdot {\mathrm{sin}}^{-1}\mathrm{x}-\frac{1}{2}{\mathrm{sin}}^{-1}\mathrm{x}+\frac{1}{2}\int \sqrt{1-{\mathrm{x}}^{2}}\mathrm{dx}\end{array}$

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