Q. If $f(x) = \tan{x - \tan^{3}{x + \tan^{5}{x - ...\infty\ with\ 0 < x < \frac{\pi}{4},\ then\ \int_{0}^{\frac{\pi}{4}}{f(x)dx}}}}$ is equal to
A) 1
B) 0
C) $\frac{1}{4}$
D) $\frac{1}{2}$
Explanation
(c)
$f(x) = \tan{x - \tan^{3}{x + \tan^{5}{x - ...\infty\ }}}$
$f(x) = \frac{\tan x}{1 + \tan^{2}x} = \frac{\tan x}{\sec^{2}x} = \frac{\sin{2x}}{2}$
$\therefore\ \int_{0}^{\frac{\pi}{4}}{f(x)dx} = \int_{0}^{\frac{\pi}{4}}{\frac{\sin{2x}}{2}dx}$
$= \left\lbrack - \frac{\cos{2x}}{4} \right\rbrack_{0}^{\frac{\pi}{4}} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.