Question

$\int \frac{\tan \left(\frac{\pi }{4}-x\right)}{{\cos }^{2}x\sqrt{{\tan }^{3}x+{\tan }^{2}x+\tan x}}dx$ 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 $\tan x=t$
$\therefore I=\int \frac{(1-t)}{(1+t)\sqrt{t^3+t^2+t}}dt=\int \frac{\left(1-t^2\right)dt}{\left(1+t^2+2t\right)\sqrt{t^3+t^2+t}}$
$=\int \frac{\left(\frac{1}{t^2}-1\right)dt}{\left(t+2+\frac{1}{t}\right)\sqrt{t+1+\frac{1}{t}}}\text{ Let }t+1+\frac{1}{t}=u^2$
$=-\int \frac{2udu}{\left(u^2+1\right)u}=-2{\tan }^{-1}\sqrt{t+1+\frac{1}{t}}+C$