Question

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

Answer 1

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

$=-\int \frac{-2-\sin 2x}{2-\sin 2x}dx$

$=-\int \frac{\left(2-\sin 2x\right)-4}{2-\sin 2x}dx$

$=-\int dx+4\int \frac{dx}{2-\sin 2x}$

$=-x+4\int \frac{dx}{2-\frac{2\tan x}{1+{\tan }^{2}x}}$

$=-x+\frac{4}{2}\int \frac{{\sec }^{2}xdx}{1+{\tan }^{2}x-\tan x}$

$Putting\tan x=t{\sec }^{2}xdx=dt$

$=-x+2\int \frac{dt}{t^2-t+1}$

$=-x+2\int \frac{dt}{t^2-t+\frac{1}{4}-\frac{1}{4}+1}$

$=-x+2\int \frac{dt}{{\left(t-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$=-x+2\times \frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+c$

$=-x+\frac{4}{\sqrt{3}}{\tan }^{-1}\left(\frac{2t-1}{\sqrt{3}}\right)+c$

$=-x+\frac{4\sqrt{3}}{3}{\tan }^{-1}\left(\frac{2\tan x-1}{\sqrt{3}}\right)+c$