Question

$\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{|x|dx}{8{\cos }^{2}2x+1}$ has the value :

• A. $\frac{{\pi }^2}{6}$
• B. $\frac{{\pi }^2}{12}$
• C. $\frac{{\pi }^2}{24}$
• D. $\frac{\pi }{12}$

Answer 1

Answer: (B)

$\underset{-\frac{\pi }{2}}{\overset{\pi /2}{\int }}\frac{|x|dx}{8{\cos }^{2}2x+1}$
$=2\underset{0}{\overset{\pi /2}{\int }}\frac{xdx}{8{\cos }^{2}2x+1}=2I$
$I=\underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{\pi }{2}-x\right)dx}{8{\cos }^2(\pi -2x)+1}$
$I=\frac{\pi }{2}\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{8{\cos }^{2}2x+1}-I=\pi \underset{0}{\overset{\pi /4}{\int }}\frac{dx}{8{\cos }^{2}2x+1}$
$\therefore I=\frac{\pi }{2}\underset{0}{\overset{\pi /4}{\int }}\frac{dx}{8{\cos }^{2}2x+1}$
$=\frac{\pi }{4}\underset{0}{\overset{\pi /4}{\int }}\frac{2{\sec }^{2}2xdx}{9+{\tan }^{2}2x}=\frac{\pi }{4}\cdot \frac{1}{3}{\tan }^{-1}{\left(\frac{1}{3}\tan 2x\right)}_0^{\pi /4}=\frac{{\pi }^2}{24}$
$\therefore$ given integral $=\frac{{\pi }^2}{12}$