Question

Find $\int \frac{dx}{\sqrt{2x-x^2}}$

Answer 1

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

Substitute $x-1=sin\theta \Rightarrow dx=cos\theta d\theta$

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

$=\int \frac{cos\theta d\theta }{cos\theta }=\int d\theta =\theta +C$ ......... [Where C is a constant]

But $\theta ={sin}^{-1}(x-1)$

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