Question

If $f\left(x\right)=\tan x-{\tan }^{3}x+{\tan }^{5}x-.........\infty$ with $0<x<\frac{\mathrm{\pi }}{4}$, then ${\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(x\right)dx=$

• A. $1$
• B. $0$
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$
• E. $-\frac{1}{4}$

Answer 1

The correct option is C

$\frac{1}{4}$

Explanation for the correct option:

Step 1: Simplify the given function:

Given that,

$f\left(x\right)=\tan x-{\tan }^{3}x+{\tan }^{5}x-.........\infty$

Since the function is in GP, so

$\begin{array}{rcl}f\left(x\right)& =& \frac{\tan x}{1-\left(-{\tan }^{2}x\right)}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{a}}{\mathbf{1}\mathbf{-}\mathbf{r}}\mathbf{]}\\ & =& \frac{\tan x}{{\sec }^{2}x}\\ & =& \frac{{\displaystyle \frac{\sin x}{\cos x}}}{{\displaystyle \frac{1}{{\cos }^{2}x}}}\\ & =& \frac{2}{2}\left(\sin x\cos x\right)\\ & =& \frac{\sin 2x}{2}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{sin}\mathbf{x}\mathbf{cos}\mathbf{x}\mathbf{]}\end{array}$

Step 2: Finding the value of ${\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(x\right)dx$:

Now, integrate the simplified function from $0$ to $\frac{\mathrm{\pi }}{4}$,

$\begin{array}{rcl}{\int }_0^{\frac{\mathrm{\pi }}{4}}f\left(x\right)dx& =& {\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{\sin 2x}{2}dx\\ & =& {\left[\frac{-\cos 2x}{4}\right]}_0^{\frac{\mathrm{\pi }}{4}}\\ & =& \frac{1}{4}\left[-\cos 2\left(\frac{\mathrm{\pi }}{4}\right)+\cos \left(0\right)\right]\\ & =& \frac{1}{4}\left[\because \cos 0°=1\right]\end{array}$

Hence, the correct option is C.