# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{\sqrt{2 x-x^{2}}} d x$

Solution:

let $I=\int \frac{1}{\sqrt{2 x-x^{2}}} d x$

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

$=\int \frac{1}{\sqrt{-\left[x^{2}-2 x(1)+1^{2}-1^{2}\right]}} d x$

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

$=\int \frac{1}{\sqrt{1-(x-1)^{2}}} d x$

let $(x-1)=t$

$d x=d t$

SO, I $=\int \frac{1}{\sqrt{1-t^{2}}} d t$

$=\sin ^{-1} \mathrm{t}+\mathrm{c}\left[\right.$ since $\left.\int \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\sin ^{-1} \mathrm{x}+\mathrm{c}\right]$

$I=\sin ^{-1}(x-1)+c$