Answer
The solution is $$\frac{2}{\sqrt5}\arctan\sqrt5.$$
Work Step by Step
To solve the integral
$$\int_{0}^{\pi/2}\frac{1}{3-2\cos\theta}d\theta$$
we will use substitution $\tan\frac{\theta}{2}=t\Rightarrow d\theta=\frac{2}{1+t^2}dt,\cos\theta=\frac{1-t^2}{1+t^2}$ and for the integration bounds we will have $t=0$ and $t=1$. Putting all of this into the integral we will have:
$$\int_{0}^{\pi/2}\frac{1}{3-2\cos\theta}d\theta=\int_{0}^{1}\frac{1}{3-2\frac{1-t^2}{1+t^2}}\frac{2}{1+t^2}dt=\int_{0}^{1}\frac{1}{\frac{3(1+t^2)-2(1-t^2)}{1+t^2}}\frac{2}{1+t^2}dt=\int_{0}^{1}\frac{1+t^2}{3+3t^2-2+2t^2}\frac{2}{1+t^2}dt=2\int_{0}^{1}\frac{dt}{1+5t^2}.$$
Now we will use new substitution $z=\sqrt5 t$ which gives us $\sqrt5 dt=dz$ and the integration bounds are $z=0$ and $z=\sqrt5$. Putting this into our integral we have:
$$2\int_{0}^{1}\frac{dt}{1+5t^2}=2\int_{0}^{\sqrt5}\frac{1}{1+z^2}\frac{dz}{\sqrt5}=\frac{2}{\sqrt5}\left.\arctan z\right|_{0}^{\sqrt5}=\frac{2}{\sqrt5}(\arctan\sqrt5-\arctan0)=\frac{2}{\sqrt5}\arctan\sqrt5-\frac{2}{\sqrt5}0=\frac{2}{\sqrt5}\arctan\sqrt5.$$
