Question

Solve:
$\int {\cos }^{-1}xdx$

Answer 1

Consider the given integral.

$I=\int {\cos }^{-1}xdx$

$I=\int 1.{\cos }^{-1}xdx$

We know that

$\int uvdx=u\int vdx-\int (\frac{d}{dx}\left(u\right)\int vdx)dx$

Therefore,

$I={\cos }^{-1}x\left(x\right)-\int (-\frac{1}{\sqrt{1-x^2}})\left(x\right)dx$

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

Let $t=1-x^2$

$\frac{dt}{dx}=0-2x$

$-\frac{dt}{2}=xdx$

Therefore,

$I=x{\cos }^{-1}x-\frac{1}{2}\int \frac{dt}{\sqrt{t}}$

$I=x{\cos }^{-1}x-\frac{1}{2}\left(2\sqrt{t}\right)+C$

$I=x{\cos }^{-1}x-\sqrt{t}+C$

On putting the value of $t$, we get

$I=x{\cos }^{-1}x-\sqrt{1-x^2}+C$

Hence, this is the answer.