# Evaluate : \begin{align*} \int \frac{tan \: x+tan^3x}{1+tan^3x} \end{align*}

Toolbox:
• $1+\tan^2x=\sec^2x$
• Form of the partial function $\large\frac{Ax+B}{(x-a)}+\frac{C}{(x-b)}$
Step 1:
$I= \Large\int\large\frac{\tan x+\tan^3xdx}{1+\tan^3x}$
$\quad= \Large\int\large\frac{\tan x(1+\tan^2x)}{1+\tan^3x}$$dx 1+\tan^2x=\sec^2x \therefore I= \Large\int\large\frac{\tan x.\sec^2x}{1+\tan^3x}$$dx$
Step 2:
Put $t=\tan x$
$dt=\sec^2xdx$
$I= \Large\int\large\frac{tdt}{1+t^3}$
$(1+t^3)=(1+t)(1-t+t^2)$
$I= \Large\int\large\frac{tdt}{(1+t)(1-t+t^2)}$
Resolving into partial fraction we get,
$\large\frac{t}{(1+t)(1-t+t^2)}=\frac{A}{(1+t)}+\frac{Bx+c}{1-t+t^2}$
$t=A(1-t+t^2)+(Bx+C)(1+t)$
Put $t=-1$
$-1=A(2+1)$
$\Rightarrow A=\large\frac{-1}{3}$
Comparing the coeff of $t^2$ on both sides we get,
$A+B=0$
$\Rightarrow B=\large\frac{1}{3}$
Comparing the constant term we get,
$A+C=0$
$C=\large\frac{1}{3}$
$\therefore \large\frac{t}{1+t^3}=\frac{1}{3(1+t)}+\frac{(1/3)t+1/3}{1-t+t^2}$
Step 3:
$I=\large\frac{-1}{3} \Large\int\large\frac{1}{t+1}$$dt+\large\frac{1}{3} \Large\int\frac{t+1}{1-t+t^2}$$dt$
Multiply and divide $I_2$ by 2
$I=\large\frac{-1}{3} \Large\int\large\frac{1}{t+1}$$dt+\large\frac{1}{6}\int\frac{2t-1+3}{1-t+t^2}$$dt$
$\;\;=-\large\frac{1}{3}$$\log\mid 1+t\mid+\large\frac{1}{6}\int\frac{2t-1}{1-t+t^2}$$dt+\large\frac{3}{6}\int\frac{1}{t^2-t+1}$$dt \;\;=-\large\frac{1}{3}$$\log\mid 1+t\mid+\large\frac{1}{6}$$\log\mid 1-t+t^2\mid+\large\frac{1}{2}.\frac{1}{(t-1/2)^2+(\sqrt 3/2)^2}$$dt$
$\;\;=-\large\frac{1}{3}$$\log\mid 1+t\mid+\large\frac{1}{6}$$\log\mid 1-t+t^2\mid+\large\frac{1}{\sqrt 3}$$\tan^{-1}\big(\large\frac{2t-1}{\sqrt 3}\big)+c Step 4: Substituting for t we get, I=-\large\frac{1}{3}$$\log\mid 1+\tan x\mid+\large\frac{1}{6}$$\log\mid 1-\tan x+\tan^2x\mid+\large\frac{1}{\sqrt 3}$$\tan^{-1}\big(\large\frac{2\tan x-1}{\sqrt 3}\big)+c$
edited Dec 1 by meena.p