Question

Solve :
$\int \frac{1}{2+\tan x}dx$

Answer 1

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

Let $u=\tan x\Rightarrow du={\sec }^{2}xdx$

$\Rightarrow du=(1+{\tan }^{2}x)dx$

$\Rightarrow du=(1+u^2)dx$

$\Rightarrow \frac{du}{(1+u^2)}=dx$

Now, $\int \frac{dx}{2+\tan x}=\int \frac{du}{(1+u^2)(2+u)}$

Consider $\frac{1}{(1+u^2)(2+u)}=\frac{A}{2+u}+\frac{Bu+c}{1+u^2}$ by the method of partial fractions

$\Rightarrow 1=A(1+u^2)+(Bu+c)(2+u)$

Put $u=-2$

$\Rightarrow 1=A(1+4)$

$\Rightarrow 5A=1$

$\therefore A=\frac{1}{5}$

Put $u=0$

$\Rightarrow 1=A+2C$

$\Rightarrow 2C=1-A=1-\frac{1}{5}=\frac{5-1}{5}=\frac{4}{5}$ since $A=\frac{1}{5}$

$\therefore C=\frac{2}{5}$

Put $u=1$

$\Rightarrow 1=2A+3B+3C$

$\Rightarrow 1=2\times \frac{1}{5}+3B+3\times \frac{2}{5}$

$\Rightarrow 1=3B+\frac{8}{5}$

$\Rightarrow 3B=1-\frac{8}{5}=\frac{5-8}{5}=\frac{-3}{5}$

$\therefore B=\frac{-1}{5}$

Hence the equation $\frac{1}{(1+u^2)(2+u)}=\frac{A}{2+u}+\frac{Bu+c}{1+u^2}$ becomes

$\frac{1}{(1+u^2)(2+u)}=\frac{1}{5(2+u)}+\frac{\frac{-3}{5}u+\frac{2}{5}}{1+u^2}$

Integrating both sides, we get

$\int \frac{du}{(1+u^2)(2+u)}=\int \frac{du}{5(2+u)}+\int \frac{(\frac{-3}{5}u+\frac{2}{5})du}{1+u^2}$

$=\frac{1}{5}\int \frac{du}{u+2}-\frac{1}{5}\left[\frac{-3}{2}\int \frac{2udu}{1+u^2}+2\int \frac{du}{1+u^2}\right]$

$=\frac{1}{5}\log |u+2|+\frac{3}{10}\log |1+u^2|+\frac{2}{5}{\tan }^{-1}u+c$

$=\frac{1}{5}\log |\tan x+2|+\frac{3}{10}\log |1+{\tan }^{2}x|+\frac{2}{5}{\tan }^{-1}\left(\tan x\right)+c$