Q. $\int_{}^{}{\frac{\log_{e}{(\tan x)}}{\sin{x\cos x}}dx}$ is equal to
A) $\left\lbrack \log_{e}\left( \tan x \right) \right\rbrack^{2} + c$
B) $\frac{1}{2}\left( \log_{e}{\tan x} \right)^{2} + c$
C) $\log_{e}{\left( \log_{e}{\tan x} \right) + c}$
D) $\log_{e}{\tan{x + c}}$
Explanation
(b)
Let $I = \int_{}^{}{\frac{\log_{e}{(\tan x)}}{\sin{x\cos x}}dx} = \int_{}^{}{\frac{\log_{e}{(\tan x)}}{\tan x}\sec^{2}xdx}$
Put $\tan{x = t\ \ \Rightarrow \ \sec^{2}xdx = dt}$
$\therefore I = \int_{}^{}\frac{\log_{e}t}{t}dt$
Again, put $\log_{e}{t = u\ \ \Rightarrow \ \frac{1}{t}}dt = du$
$\therefore\ I = \int_{}^{}{u\ du} = \frac{u^{2}}{2} + c$
$= \frac{\left( \log_{e}t \right)^{2}}{2} + c$
$= \frac{1}{2}{\lbrack\log_{e}{\tan x}\rbrack}^{2} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.