x=2arctgt;dx=\frac{2dt}{1+t^2}\\cosx=\frac{1-t^2}{1+t^2};2cosx+3=\frac{t^2+5}{1+t^2}" alt="\displaystyle \int\limits^\pi_0\frac{dx}{3+2cosx}=\int\limits^\pi_0\frac{2dt}{1+t^2}*\frac{1+t^2}{t^2+5}=2\sqrt5\int\limits^\infty_0\frac{d\frac{t}{\sqrt5}}{\frac{t^2}{5}+1}= \lim_{b\to\infty}\frac{2}{\sqrt5}arctg\frac{t}{\sqrt{5}}|^b_0=\\=\frac{2}{\sqrt5}\lim_{b\to\infty}(arctg\frac{b}{\sqrt5})=\frac{\pi}{\sqrt5}\\t=tg\frac{x}{2}=>x=2arctgt;dx=\frac{2dt}{1+t^2}\\cosx=\frac{1-t^2}{1+t^2};2cosx+3=\frac{t^2+5}{1+t^2}" align="absmiddle" class="latex-formula">