Question

$\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\cos }^{2}2x}{1+{25}^x}dx=$______

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

Answer 1

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

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\cos }^{2}2x}{1+{25}^x}dx$

$I=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{{\cos }^{2}2x}{1+{25}^{-x}}dx$
Adding these two equations
$I=\frac{1}{2}\underset{-\pi /2}{\overset{\pi /2}{\int }}(\frac{{\cos }^{2}2x}{1+{25}^x}+\frac{{\cos }^{2}2x}{1+{25}^{-x}})dx$

$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(\frac{{\cos }^{2}2x}{1+{25}^x}+\frac{{\cos }^{2}2x}{1+{25}^{-x}})dx$

$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(\frac{{\cos }^{2}2x}{1+{25}^x}+\frac{{25}^x{\cos }^{2}2x}{1+{25}^x})dx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}2xdx$
$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}\frac{1+\cos 4x}{2}dx$
$=\frac{1}{2}{[x+\frac{\sin 4x}{4}]}_0^{\pi /2}$

$=\frac{1}{2}[\frac{\pi }{2}+0-0+0]$

$=\frac{\pi }{4}$