Question

If $f(x)=\tan x-{\tan }^{3}x+{\tan }^{5}x-.....$ to $\infty$ 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. $\frac{1}{2}$
• E. $-\frac{1}{4}$

Answer 1

Answer: (C)

$f(x)=\tan x-{\tan }^{3}x+{\tan }^{5}x-....\infty$ $f(x)=\frac{\tan x}{1+{\tan }^{2}x}=\frac{\tan x}{{\sec }^{2}x}=\frac{\sin 2x}{2}$ $\therefore$ ${\int }_0^{\pi /4}f(x)dx={\int }_0^{\pi /4}\frac{\sin 2x}{2}$ $={\left[-\frac{\cos 2x}{4}\right]}_0^{\pi /4}$ $=\frac{1}{4}$