Question

The value of the integral $\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{x+\frac{\pi }{4}}{2-\cos 2x}dx$ 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)

$I=\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{x+\frac{\pi }{4}}{2-\cos 2x}dx$ (1)
$x\to -x$
$I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{-x+\frac{\pi }{4}}{2-\cos 2x}dx$ (2)
$(1)+(2)$
$2I=\underset{\frac{-\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}\frac{\frac{\pi }{2}}{2-\cos 2x}dx$
$I=\frac{\pi }{4}\cdot 2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{2-\cos 2x}dx$
$I=\frac{\pi }{4}\cdot 2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\left(1+{\tan }^{2}x\right)dx}{2\left(1+{\tan }^{2}x\right)-\left(1-{\tan }^{2}x\right)}$
$I=\frac{\pi }{4}\underset{0}{\overset{1}{\int }}\frac{dt}{3t^2+1}$
$\Rightarrow I=\frac{\pi }{2\sqrt{3}}{\tan }^{-1}\sqrt{3}$
$I=\frac{{\pi }^2}{6\sqrt{3}}$