Question

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

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

Answer 1

Answer: (A)

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\cos }^{2}x}{1+3^x}dx$ (using king)
$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\cos }^{2}x}{1+3^{-x}}dx=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{3^x{\cos }^{2}x}{1+3^x}dx$
$2I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{\left(1+3^x\right){\cos }^{2}x}{1+3^x}dx$
$=\underset{-\pi /2}{\overset{\pi /2}{\int }}{\cos }^{2}xdx=2\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}xdx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}xdx=\frac{\pi }{4}$