Este scrisa mai pe scurt rezolvarea, sper sa o intelegi.
[tex] \\ \int ^{\frac{\Pi }{2}}_0\:\frac{dx}{2cosx+3} =>
\\
\\ =\int \frac{2}{u^2+5}du
\\
\\ =2\cdot \int \:\frac{1}{u^2+5}du
\\
\\=2\cdot \frac{1}{\sqrt{5}}\cdot \int \:\frac{1}{v^2+1}dv
\\
\\ =2\cdot \frac{1}{\sqrt{5}}\arctan \left(v\right)
\\
\\ =2\cdot \frac{1}{\sqrt{5}}\arctan \left(\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{5}}\right)
\\ =\frac{2}{\sqrt{5}}\arctan \left(\frac{1}{\sqrt{5}}\right)-0
\\
\\ \boxed{=\frac{2}{\sqrt{5}}\arctan \left(\frac{1}{\sqrt{5}}\right)} [/tex]
