Question

The value of $\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\cos }^{2}x}{1+3^x}dx$ is :

• A. $2\pi$
• B. $4\pi$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

Let $I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\cos }^{2}x}{1+3^x}dx\dots \left(1\right)$
$\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\cos }^2(0-x)}{1+3^{0-x}}dx$
$\Rightarrow I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{3^x{\cos }^{2}x}{1+3^x}dx\dots \left(2\right)$

From $\left(1\right)+\left(2\right),$ we get
$2I=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\cos }^{2}xdx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}xdx=\frac{\pi }{4}$