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Question

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

A

$2\mathrm{\pi }$

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B

$\frac{\mathrm{\pi }}{a}$

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C

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

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D

$a\mathrm{\pi }$

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Solution

## The correct option is C $\frac{\mathrm{\pi }}{2}$Explanation for the correct option:Compute the required value:Given: ${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\mathrm{cos}}^{2}\left(x\right)}{1+{a}^{x}}dx$Let $I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\mathrm{cos}}^{2}\left(x\right)}{1+{a}^{x}}dx-------\left(1\right)$substitute $\mathrm{x}\to \mathrm{\pi }+\left(-\mathrm{\pi }\right)-\mathrm{x}⇒-\mathrm{x}$$I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\mathrm{cos}}^{2}\left(x\right)}{1+{a}^{x}}dx⇒{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\mathrm{cos}}^{2}\left(-x\right)}{1+{a}^{-x}}dx\phantom{\rule{0ex}{0ex}}I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\mathrm{cos}}^{2}\left(x\right)}{1+\frac{1}{{a}^{x}}}dx\phantom{\rule{0ex}{0ex}}I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{a}^{x}{\mathrm{cos}}^{2}\left(x\right)}{{a}^{x}+1}dx-----\left(2\right)$Add equations $\left(1\right)$ and $\left(2\right)$$I+I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{\mathrm{cos}}^{2}\left(x\right)}{1+{a}^{x}}dx+{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{{a}^{x}{\mathrm{cos}}^{2}\left(x\right)}{1+{a}^{x}}dx\phantom{\rule{0ex}{0ex}}2I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{\left(1+{a}^{x}\right){\mathrm{cos}}^{2}\left(x\right)}{1+{a}^{x}}dx\phantom{\rule{0ex}{0ex}}2I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\mathrm{cos}}^{2}\left(x\right)dx\phantom{\rule{0ex}{0ex}}2I={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\frac{1+\mathrm{cos}2x}{2}dx\phantom{\rule{0ex}{0ex}}I=\frac{1}{4}\left[{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left(1\right)·dx+{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\mathrm{cos}\left(2x\right)dx\right]\phantom{\rule{0ex}{0ex}}I=\frac{1}{4}\left[{\left|x\right|}_{-\mathrm{\pi }}^{\mathrm{\pi }}+{\left|\frac{\mathrm{sin}\left(2x}{2}\right|}_{-\mathrm{\pi }}^{\mathrm{\pi }}\right]\phantom{\rule{0ex}{0ex}}I=\frac{1}{4}\left[\mathrm{\pi }+\mathrm{\pi }+\frac{\mathrm{sin}\left(2\mathrm{\pi }\right)+\mathrm{sin}\left(2\mathrm{\pi }\right)}{2}\right]\phantom{\rule{0ex}{0ex}}I=\frac{\mathrm{\pi }}{2}$Hence, option $A$ is the correct answer.

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