Hi there, I have to test the convergence of this integral

Question

The integral is

$ \int_{0}^{\pi/2}{\frac{1}{2+tan(x)}dx}$

thank you very much!

Answer 1

Let $u=\tan x$ and $$dx=\frac{du}{u^2+1}$$ So $$\int_0^{\pi/2} \frac{dx}{2+\tan x} = \int_0^{\infty} \frac{du}{(u+2)(u^2+1)}$$ Now, use partial fractions and we get $$= \int_0^{\infty} \left( \frac{1}{5(u+2)}-\frac{u-2}{5(u^2+1)} \right) du = \left( \frac{1}{5}\log|u+2| -\frac{1}{10} \log(u^2+1)+\frac{2}{5}\arctan(u) \right) \bigg|_0^{\infty} = \frac{\pi-\log 2}{5}$$ Hence, the integral converge.