Question

$\int\limits_{-\pi / 2}^{\pi / 2} \frac{|x| d x}{8 \cos ^2 2 x+1}$ 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)

$\int\limits_{-\frac{\pi}{2}}^{\pi / 2} \frac{|x| d x}{8 \cos ^2 2 x+1}$
$=2 \int\limits_0^{\pi / 2} \frac{x d x}{8 \cos ^2 2 x+1}=2 I$
$I=\int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) d x}{8 \cos ^2(\pi-2 x)+1}$
$I=\frac{\pi}{2} \int\limits_0^{\pi / 2} \frac{d x}{8 \cos ^2 2 x+1}-I=\pi \int\limits_0^{\pi / 4} \frac{d x}{8 \cos ^2 2 x+1} $
$\therefore I=\frac{\pi}{2} \int\limits_0^{\pi / 4} \frac{d x}{8 \cos ^2 2 x+1}$
$=\frac{\pi}{4} \int\limits_0^{\pi / 4} \frac{2 \sec ^2 2 x d x}{9+\tan ^2 2 x}=\frac{\pi}{4} \cdot \frac{1}{3} \tan ^{-1}\left(\frac{1}{3} \tan 2 x\right)_0^{\pi / 4}=\frac{\pi^2}{24}$
$\therefore $ given integral $=\frac{\pi^2}{12}$