Question

Value of $\int \frac{dx}{x^2+2x\cos \alpha +1}$ equal to

Answer 1

$I=\int \frac{dx}{(x^2+2x\cos \alpha +1)}$

$=\int \frac{dx}{(x^2+2\cos \alpha +{\cos }^2\alpha +{\sin }^2\alpha )}$

$=\int \frac{dx}{(x+\cos \alpha {)}^2+(\sin \alpha {)}^2}$

$=\frac{1}{\sin \alpha }{\tan }^{-1}\frac{x+\cos \alpha }{\sin \alpha }+c$