Question

Evaluate each of the following integrals:

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

Answer 1

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

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\frac{{\cos }^{2}x}{1+e^x}+\frac{e^x{\cos }^{2}x}{1+e^x}\right)dx \\ \Rightarrow 2I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^{2}x\left(1+e^x\right)}{1+e^x}dx \\ \Rightarrow 2I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}{\cos }^{2}xdx \\ \Rightarrow 2I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\frac{1+\cos 2x}{2}\right)dx \end{gathered}$$
$$\begin{gathered} \Rightarrow 2I=\frac{1}{2}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}dx+\frac{1}{2}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\cos 2xdx \\ \Rightarrow 2I=\frac{1}{2}\times {\overline{)x}}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}+\frac{1}{2}\times {\overline{)\frac{\sin 2x}{2}}}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow 2I=\frac{1}{2}\left[\frac{\mathrm{\pi }}{2}-\left(-\frac{\mathrm{\pi }}{2}\right)\right]+\frac{1}{4}\left[\mathrm{sin\pi }-\sin \left(-\mathrm{\pi }\right)\right] \\ \Rightarrow 2I=\frac{1}{2}\times \mathrm{\pi }+\frac{1}{4}\left(0+0\right)\left[\sin \left(-\mathrm{\pi }\right)=-\mathrm{sin\pi }=0\right] \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{2} \\ \Rightarrow I=\frac{\mathrm{\pi }}{4} \end{gathered}$$