Question

$\int \frac{2}{2+\sin 2x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{2}{2+\sin \left(2x\right)}dx \\ =\int \frac{2}{2+2\sin x\cos x}dx \\ =\int \frac{1}{1+\sin x\cos x}dx \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{\cos }^{2}x \\ \Rightarrow I=\int \frac{{\sec }^{2}xdx}{{\sec }^{2}x+\tan x} \\ =\int \frac{{\sec }^{2}xdx}{1+{\tan }^{2}x+\tan x} \\ \mathrm{Let}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int \frac{dt}{t^2+t+1} \\ =\int \frac{dt}{t^2+t+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}+1} \\ =\int \frac{dt}{{\left(t+{\displaystyle \frac{1}{2}}\right)}^2+{\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2} \\ =\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{t+{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)+C \\ =\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t+1}{\sqrt{3}}\right)+C \\ =\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x+1}{\sqrt{3}}\right)+C \end{gathered}$$