Question

The value of the integral $\int x{\sin }^{-1}xdx$ is equal to

• A. $\frac{1}{2}{x}^2{\sin }^{-1}x+\frac{1}{4}x\sqrt{1-x^2}-\frac{1}{4}{\sin }^{-1}x+C$
• B. $\frac{1}{2}{x}^2{\sin }^{-1}x-\frac{1}{4}x\sqrt{1-x^2}-\frac{1}{4}{\sin }^{-1}x+C$
• C. $\frac{1}{2}{x}^2{\sin }^{-1}x+\frac{1}{4}x\sqrt{1-x^2}+\frac{1}{4}{\sin }^{-1}x+C$
• D. $\frac{1}{2}{x}^2{\sin }^{-1}x+\frac{1}{4}\sqrt{1-x^2}-\frac{1}{4}{\sin }^{-1}x+C$

Answer 1

Answer: (A)

Let $I=\int x{\sin }^{-1}xdx$
$\Rightarrow I=({\sin }^{-1}x\frac{x^2}{2}-\int \frac{1}{\sqrt{1-x^2}}\cdot \frac{x^2}{2}dx$
$\Rightarrow I=\frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\int \frac{-x^2}{\sqrt{1-x^2}}dx$
$=\frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\int \frac{1-x^2-1}{\sqrt{1-x^2}}dx$
$\Rightarrow =\frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\left\{\int \frac{1-x^2}{\sqrt{1-x^2}}dx-\int \frac{1}{\sqrt{1-x^2}}dx\right\}$
$\Rightarrow I=\frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\left\{\int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx\right\}$
$=\frac{x^2}{2}{\sin }^{-1}x+\frac{1}{2}\left[\left\{\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x\right\}-{\sin }^{-1}x\right]+C$
$\Rightarrow I=\frac{1}{2}{x}^2{\sin }^{-1}x+\frac{1}{4}x\sqrt{1-x^2}-\frac{1}{4}{\sin }^{-1}x+C$