Question

Evaluate each of the following integrals:

${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\tan }^{2}x}{1+e^x}dx$

Answer 1

Let I = ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\tan }^{2}x}{1+e^x}dx$ .....(1)
Then,
$$\begin{gathered} I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\tan }^2\left[{\displaystyle \frac{\mathrm{\pi }}{4}}+\left(-{\displaystyle \frac{\mathrm{\pi }}{4}}\right)-x\right]}{1+e^{\left[{\displaystyle \frac{\mathrm{\pi }}{4}}+\left(-{\displaystyle \frac{\mathrm{\pi }}{4}}\right)-x\right]}}dx\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right] \\ ={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\tan }^2\left(-x\right)}{1+e^{-x}}dx \\ ={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{{\tan }^{2}x}{1+{\displaystyle \frac{1}{e^x}}}dx \\ ={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{e^x{\tan }^{2}x}{e^x+1}dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\left(\frac{{\tan }^{2}x}{1+e^x}+\frac{e^x{\tan }^{2}x}{1+e^x}\right)dx \\ \Rightarrow 2I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{\left(1+e^x\right){\tan }^{2}x}{1+e^x}dx \\ \Rightarrow 2I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\tan }^{2}xdx \\ \Rightarrow 2I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\left({\sec }^{2}x-1\right)dx \end{gathered}$$
$$\begin{gathered} \Rightarrow 2I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\sec }^{2}xdx-{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}dx \\ \Rightarrow 2I={\overline{)\tan x}}_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}-{\overline{)x}}_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}} \\ \Rightarrow 2I=\left[\tan \frac{\mathrm{\pi }}{4}-\tan \left(-\frac{\mathrm{\pi }}{4}\right)\right]-\left[\frac{\mathrm{\pi }}{4}-\left(-\frac{\mathrm{\pi }}{4}\right)\right] \\ \Rightarrow 2I=\left(1+1\right)-\left(\frac{2\mathrm{\pi }}{4}\right) \\ \Rightarrow 2I=2-\frac{\mathrm{\pi }}{2} \\ \Rightarrow I=1-\frac{\mathrm{\pi }}{4} \end{gathered}$$


Disclaimer: This answer does not matches with the given answer in the book.