How to integrate $\frac{1}{\sin(x)+2}$

Question

I want to find the indefinite integral of $\frac{1}{\sin(x)+2}$. I know that the integral of $\frac{1}{\sin(x)}$ can be easily obtained by $x:=2u$, and the final solution is $\ln(\tan(x/2))$. However, I do not know how to handle the shift in the denominator. Thank you, in advance, for your help.

Answer 1

The tan half substitution solves this

$$\int\frac{1}{\sin x + 2}\mathrm{d}x \underset{u=\tan\frac{x}{2}}= \int\frac{2}{(u^2+1)(\frac{2u}{u^2+1}+2)}\mathrm{d}u = \int\frac{1}{1+u+u^2}\mathrm{d}u$$

Finishing from here should not be that hard.