Question

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

Answer 1

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

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

Put $(x-1)=t$

$dx=dt$.

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

Put $t=\sin \theta$

$dt=\cos \theta d\theta$

$I=\int \frac{\cos \theta d\theta }{\sqrt{1-{\sin }^2\theta }}=\int \frac{\cos \theta d\theta }{\cos \theta }=\int d\theta$

$I=\theta +c={\sin }^{-1}\left(t\right)+c={\sin }^{-1}(x-1)+c$

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