Question

$\int \frac{\tan \left(\frac{\pi}{4}-x\right)}{\cos ^2 x \sqrt{\tan ^3 x+\tan ^2 x+\tan x}} d x$ is equal to

• A. $-2 \tan ^{-1} \sqrt{\tan x+1+\frac{1}{\tan x}}+C$
• B. $2 \tan ^{-1} \sqrt{\tan x+1+\frac{1}{\tan x}}+C$
• C. $-3 \tan ^{-1} \sqrt{\tan x+1+\frac{1}{\tan x}}+C$
• D. $3 \tan ^{-1} \sqrt{\tan x+1+\frac{1}{\tan x}}+C$ (where $C$ is constant of integration.)

Answer 1

Answer: (A)

Let $\tan x=t$
$\therefore I =\int \frac{(1-t)}{(1+t) \sqrt{t^3+t^2+t}} d t=\int \frac{\left(1-t^2\right) d t}{\left(1+t^2+2 t\right) \sqrt{t^3+t^2+t}} $
$ =\int \frac{\left(\frac{1}{t^2}-1\right) d t}{\left(t+2+\frac{1}{t}\right) \sqrt{t+1+\frac{1}{t}}} \text { Let } t+1+\frac{1}{t}=u^2 $
$ =-\int \frac{2 u d u}{\left(u^2+1\right) u}=-2 \tan ^{-1} \sqrt{t+1+\frac{1}{t}}+C$