Question

Evaluate:

$I=\int \frac{1}{2+sinx+cosx}dx$

Answer 1

$\begin{array}{c}{\int }^{}\frac{1}{\sin \left(x\right)+\cos \left(x\right)+2}1mudx={\int }^{}\frac{1}{\frac{1-{\tan }^2\left(\frac{x}{2}\right)}{{\tan }^2\left(\frac{x}{2}\right)+1}+\frac{2\tan \left(\frac{x}{2}\right)}{{\tan }^2\left(\frac{x}{2}\right)+1}+2}dx\\ Substituteu=\tan \left(\frac{x}{2}\right),\\ Therefore,dx=\frac{2}{{\sec }^2\left(\frac{x}{2}\right)}1mudu\\ ={\int }^{}\frac{1}{{\left(u+1\right)}^2+2}du\\ =\sqrt{2}\tan \left(\frac{u+1}{\sqrt{2}}\right)+C\end{array}$