Question

The value of the integral $\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ is :

• A. $\frac{\pi^2}{12 \sqrt{3}}$
• B. $\frac{\pi^2}{6}$
• C. $\frac{\pi^2}{3 \sqrt{3}}$
• D. $\frac{\pi^2}{6 \sqrt{3}}$

Answer 1

Answer: (D)

$I=\int\limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ (1)
$x \rightarrow- x$
$I=\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x+\frac{\pi}{4}}{2-\cos 2 x} d x$ (2)
$(1) +(2)$
$2 I =\int\limits_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{\pi}{2}}{2-\cos 2 x} d x$
$I=\frac{\pi}{4} \cdot 2 \int\limits_0^{\frac{\pi}{4}} \frac{ dx }{2-\cos 2 x} d x$
$ I =\frac{\pi}{4} \cdot 2 \int\limits_0^{\frac{\pi}{4}} \frac{\left(1+\tan ^2 x \right) dx }{2\left(1+\tan ^2 x \right)-\left(1-\tan ^2 x \right)} $
$ I =\frac{\pi}{4} \int\limits_0^1 \frac{ dt }{3 t ^2+1} $
$ \Rightarrow I =\frac{\pi}{2 \sqrt{3}} \tan ^{-1} \sqrt{3} $
$ I =\frac{\pi^2}{6 \sqrt{3}}$