Question

Evaluate $\rm \displaystyle\int \Big(\frac{1-\tan x}{1+\tan x}\Big)^{2} \ dx$.

Answer 1

Before integration ,first convert the expression by substitution Method

$\int \Big(\frac{1-\tan x}{1+\tan x}\Big)^{2} \ dx \\ \\ \int \frac{{sec}^{2}x}{{sec}^{2}x} \Big(\frac{1-\tan x}{1+\tan x}\Big)^{2} \ dx \\ \\ \int \frac{{sec}^{2}x}{1 + {tan}^{2}x} \Big(\frac{1-\tan x}{1+\tan x}\Big)^{2} \ dx \\ \\ substitute \: tan \: x = t \\ \\ {sec}^{2} x \: dx = dt \\ \\ \int \frac{1}{1 + {t}^{2}} \Big(\frac{1-t}{1+t}\Big)^{2} \ dt \\ \\ now \: by \: partial \: fraction \: decomposition \\ \\ = \int \frac{2}{ {(t + 1)}^{2} } \ dt - \int \: \frac{1}{ {t}^{2} + 1}dt \\ \\ = 2\int \: {(t + 1)}^{ - 2} dt - \int \: \frac{1}{ {t}^{2} + 1}dt \\ \\ = - 2\: {(t + 1)}^{ (- 2 + 1)} - \: {tan}^{ - 1} t + C\\ \\ = \frac{ - 2}{t + 1} - \: {tan}^{ - 1} t + C$
Redo substitution

$= \frac{ - 2}{tan \: x + 1} - {tan}^{ - 1} (tan \: x) + C \\ \\ \int \Big(\frac{1-\tan x}{1+\tan x}\Big)^{2} \ dx \: = \frac{ - 2}{tan \: x + 1} - x + C$
Hope it helps you.