Question

$\int \frac{\ln \left(\tan x\right)}{\sin x\cos x}dx$ is equal to

• A. $\frac{1}{2}\ln \left(\tan x\right)+c$
• B. $\frac{1}{2}\ln \left({\tan }^{2}x\right)+c$
• C. $\frac{1}{2}\left(\ln \right(\tan x){)}^2+c$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2}\left(\ln \right(\tan x){)}^2+c$

$I=\int \frac{\ln \left(\tan x\right)}{\sin x\cos x}dx,$

let $t=\ln \left(\tan x\right)$
$\Rightarrow \frac{dt}{dx}=\frac{{\sec }^{2}x}{\tan x}$
$\Rightarrow dt=\frac{dx}{\sin x\cos x}$
$\Rightarrow I=\int tdt=\frac{1}{2}{t}^2+C=\frac{1}{2}\left(\ln \right(\tan x){)}^2+C$