Question

$\int \frac{si{n}^{-1}x}{\sqrt{1-x^2}}dx=$

Answer 1

Consider $\int \frac{si{n}^{-1}x}{\sqrt{1-x^2}}dx$

Let $u=si{n}^{-1}x$

$\Rightarrow du=\frac{1}{\sqrt{1-x^2}}dx$

$\therefore \int \frac{si{n}^{-1}x}{\sqrt{1-x^2}}dx=\int udu$

$=\frac{u^2}{2}+c$ where $c$ is integration constant.

Substitute back $u=si{n}^{-1}x$ we get

$\therefore \int \frac{si{n}^{-1}x}{\sqrt{1-x^2}}dx=\frac{1}{2}{\left(si{n}^{-1}x\right)}^2+c$