Question

Integration of ${\sin }^{-1}x$

Answer 1

Step 1: Separate the expression

The given expression is${\sin }^{-1}x$.

Let its integration be$I$.

$\therefore$$I=\int {\sin }^{-1}xdx$

$=\int {\sin }^{-1}x.1.dx$

Step 2: Apply Integration By parts

Let $f\left(x\right)={\sin }^{-1}x$ and$g\left(x\right)=1$.

Using integration by parts where $\int u.vdx=u\int vdx-\int u\text{'}\left(\int vdx\right)dx$ we get,

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

$=x{\sin }^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx$ $\left[\because \frac{d}{dx}\left(si{n}^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}\right]$

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

$=x{\sin }^{-1}x+\frac{1}{2}\int \left(-2x\right){\left(1-x^2\right)}^{-\frac{1}{2}}dx$

Step 3: Apply Identities of integration

We know that $\int f{\left(x\right)}^{n}f\text{'}\left(x\right)dx=\frac{f{\left(x\right)}^{n+1}}{n+1}$

$\therefore I=x{\sin }^{-1}x+\frac{1}{2}\frac{{\left(1-x^2\right)}^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}+c$ where $c$ is the constant of integration.

$=x{\sin }^{-1}x+\frac{1}{2}\frac{{\left(1-x^2\right)}^{{\displaystyle \frac{1}{2}}}}{{\displaystyle \frac{1}{2}}}+c$

$=x{\sin }^{-1}x+\sqrt{1-x^2}+c$

Hence, when ${\sin }^{-1}x$ is integrated we get $x{\sin }^{-1}x+\sqrt{1-x^2}+c$.