Question

$\int xsi{n}^{-1}xdx=$

• A. $(\frac{x^2}{2}-\frac{1}{4})si{n}^{-1}x+\frac{x}{4}\sqrt{1-x^2}+c$
• B. $(\frac{x^2}{2}+\frac{1}{4})si{n}^{-1}x+\frac{x}{4}\sqrt{1-x^2}+c$
• C. $(\frac{x^2}{2}-\frac{1}{4})si{n}^{-1}x-\frac{x}{4}\sqrt{1-x^2}+c$
• D. $(\frac{x^2}{2}+\frac{1}{4})si{n}^{-1}x-\frac{x}{4}\sqrt{1-x^2}+c$

Answer 1

The correct option is A $(\frac{x^2}{2}-\frac{1}{4})si{n}^{-1}x+\frac{x}{4}\sqrt{1-x^2}+c$

$\int xsi{n}^{-1}xdx=\frac{x^2}{2}si{n}^{-1}x-\int \frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2}dx+c=\frac{x^2}{2}si{n}^{-1}x-\frac{1}{2}\int -\frac{(1-x^2)-1}{\sqrt{1-x^2}}dx+c=\frac{x^2}{2}si{n}^{-1}x+\frac{1}{2}\int \sqrt{1-x^2}dx-\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}dx+c=\frac{x^2}{2}si{n}^{-1}x+\frac{x}{4}\sqrt{1-x^2}+\frac{1}{4}si{n}^{-1}x-\frac{1}{2}si{n}^{-1}x+c=\frac{x^2}{2}si{n}^{-1}x+\frac{x}{4}\sqrt{1-x^2}-\frac{1}{4}si{n}^{-1}x=(\frac{x^2}{2}-\frac{1}{4})si{n}^{-1}x+\frac{x}{4}\sqrt{1-x^2}+c$