Question

The value of ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\cos }^2\left(x\right)}{1+a^x}dx,a>0$ is

• A. $2\mathrm{\pi }$
• B. $\frac{\mathrm{\pi }}{a}$
• C. $\frac{\mathrm{\pi }}{2}$
• D. $a\mathrm{\pi }$

Answer 1

The correct option is C

$\frac{\mathrm{\pi }}{2}$

Explanation for the correct option:

Compute the required value:

Given: ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\cos }^2\left(x\right)}{1+a^x}dx$

Let $I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\cos }^2\left(x\right)}{1+a^x}dx-------\left(1\right)$

substitute $\mathrm{x}\to \mathrm{\pi }+(-\mathrm{\pi })-\mathrm{x}\Rightarrow -\mathrm{x}$

$$\begin{gathered} I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\cos }^2\left(x\right)}{1+a^x}dx\Rightarrow {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\cos }^2(-x)}{1+a^{-x}}dx \\ I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\cos }^2\left(x\right)}{1+{\displaystyle \frac{1}{a^x}}}dx \\ I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{a^x{\cos }^2\left(x\right)}{a^x+{\displaystyle 1}}dx-----\left(2\right) \end{gathered}$$

Add equations $\left(1\right)$ and $\left(2\right)$

$$\begin{gathered} I+I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\cos }^2\left(x\right)}{1+a^x}dx+{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{a^x{\cos }^2\left(x\right)}{1+a^x}dx \\ 2I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{(1+a^x){\cos }^2\left(x\right)}{1+a^x}dx \\ 2I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\cos }^2\left(x\right)dx \\ 2I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{1+\cos 2x}{2}dx \\ I=\frac{1}{4}\left[{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(1\right)·dx+{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\cos \left(2x\right)dx\right] \\ I=\frac{1}{4}\left[{\left|x\right|}_{-\mathrm{\pi }}^{\mathrm{\pi }}+{\left|\frac{\sin (2x}{2}\right|}_{-\mathrm{\pi }}^{\mathrm{\pi }}\right] \\ I=\frac{1}{4}\left[\mathrm{\pi }+\mathrm{\pi }+\frac{\sin \left(2\mathrm{\pi }\right)+\sin \left(2\mathrm{\pi }\right)}{2}\right] \\ I=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence, option $A$ is the correct answer.