Question

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

Answer 1

Consider the given integral.

$I=\int \frac{dx}{\sqrt{2ax-x^2}}$

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

Let $t=x-a$

$dt=dx$

Therefore,

$I=\int \frac{dt}{\sqrt{a^2-t^2}}$

We know that

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

Therefore,

$I={\sin }^{-1}\left(\frac{t}{a}\right)+C$

$I={\sin }^{-1}\left(\frac{x-a}{a}\right)+C$

Hence, this is answer.