Question

Prove the integration formula.

$\int {\sin }^{-1}xdx=x{\sin }^{-1}x+\sqrt{1-x^2}+C$

(a) by applying integration by parts to $\int {\sin }^{-1}xdx$.

(b) by differentiating$\sqrt{1-x^2}+x{\sin }^{-1}x$.

Short answer

Part (a). The solution is $x{\sin }^{-1}x+\sqrt{1-x^2}+C$.

Part (b). The solution is${\sin }^{-1}x$.

Step-by-step solution

Part (a) Step 1. Given information.

The given formula is:

$\int {\sin }^{-1}xdx=x{\sin }^{-1}x+\sqrt{1-x^2}+C$.

Part (a) Step 2. Prove the formula by applying the integration by parts formula.

Take $u={\sin }^{-1}x$and $v=1$.

role="math" localid="1650366363299" $\begin{array}{rcl}\int {\sin }^{-1}x·1dx& =& {\sin }^{-1}x\int dx-\int \left[\frac{d}{dx}\left({\sin }^{-1}x\right)\int dx\right]dx\\ & =& {\sin }^{-1}x·x-\int \frac{1}{\sqrt{1-x^2}}·xdx\\ & =& x{\sin }^{-1}x-\int \frac{1}{\sqrt{t}}\left(\frac{-dt}{2}\right)\dots \dots \dots \dots \left\{put1-x^2=t,-2xdx=dt,xdx=-\frac{dt}{2}\right\}\\ & =& x{\sin }^{-1}x+\frac{1}{2}\int \frac{1}{\sqrt{t}}dt\\ & =& x{\sin }^{-1}x+\frac{1}{2}\int t^{-\frac{1}{2}}dt\\ & =& x{\sin }^{-1}x+\frac{1}{2}\frac{t^{-{\displaystyle \frac{1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\\ & =& x{\sin }^{-1}x+\frac{1}{2}·2·\sqrt{t}+C\\ & =& x{\sin }^{-1}x+\sqrt{t}+C\end{array}$

Substitute$t=\sqrt{1-x^2}$, in the above result.

Hence, the formula$x{\sin }^{-1}x+\sqrt{1-x^2}+C$ is proved.

Part (b) Step 1. Given information.

The given formula is:

$\int {\sin }^{-1}xdx=x{\sin }^{-1}x+\sqrt{1-x^2}+C$

Part (b) Step 2. Prove the formula by differentiating x·sin-1x+1-x2.

The differentiation is calculated below:

$\begin{array}{rcl}\frac{d}{dx}\left[x·{\sin }^{-1}x+\sqrt{1-x^2}\right]& =& \frac{d}{dx}\left\{x{\sin }^{-1}x\right\}+\frac{d}{dx}\left\{\sqrt{1-x^2}\right\}\\ & =& x·\frac{d}{dx}\left({\sin }^{-1}x\right)+{\sin }^{-1}x·\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left\{\sqrt{1-x^2}\right\}\\ & =& \frac{x}{\sqrt{1-x^2}}+{\sin }^{-1}x+\frac{1}{2\sqrt{1-x^2}}\left(-2x\right)\\ & =& \frac{x}{\sqrt{1-x^2}}+{\sin }^{-1}x-\frac{x}{\sqrt{1-x^2}}\\ & =& {\sin }^{-1}x\end{array}$

Hence, the formula is proved.