Question

If $I={\int }_{-\pi /4}^{\pi /4}\frac{{\tan }^{2}x}{1+a^x}dx,a>0$, then $I$ equals

• A. $\frac{4+\pi }{4}$
• B. $\frac{4-\pi }{4}$
• C. $\frac{a\pi }{4}$
• D. $\frac{\pi +a}{4}$

Answer 1

The correct option is B $\frac{4-\pi }{4}$

Let $I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{{\tan }^{2}x}{1+a^x}dx$ ...(1)
$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{{\tan }^2(+\frac{\pi }{4}-\frac{\pi }{4}-x)}{1+a^{(\frac{\pi }{4}-\frac{\pi }{4}-x)}}dx$
$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{{\tan }^{2}x}{1+a^{-x}}dx={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{a}^x\frac{{\tan }^{2}x}{1+a^x}dx$ ...(2)
From (1) and (2), we get
$2I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\left(\frac{1+a^x}{1+a^x}\right){\tan }^{2}xdx={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}{\tan }^{2}xdx$
$={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}({\sec }^{2}x-1)dx={(\tan x-x)}_{-\frac{\pi }{4}}^{+\frac{\pi }{4}}$
Therefore $I=\frac{4-\pi }{4}$