The value of the integral _-/4^/4(x+/4)/(2-2x),dx is:

Question

The value of the integral $\displaystyle\int_{-\pi/4}^{\pi/4}\frac{x+\pi/4}{2-\cos2x}\,dx$ is:

Accepted Answer

$$
I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{(2-\cos 2 x)} d x
$$

Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$$
I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{-x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x
$$

$\begin{aligned} & \therefore 2 I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi d x}{2(2-\cos 2 x)} \\ & \Rightarrow  I=\frac{2 \pi}{4} \int_0^{\frac{\pi}{4}}\left(\frac{d x}{\left.\frac{2-1-\tan ^2 x}{1+\tan ^2 x}\right)}\right. \\ & \Rightarrow  I=\frac{\pi}{2} \int_0^{\frac{\pi}{4}}\left(\frac{1+\tan ^2 x}{1+3 \tan ^2 x}\right) d x\end{aligned}$

Now,

$$
\begin{aligned}
& \tan x=t \\
& =\frac{\pi}{2} \int_0^1 \frac{d t}{1+3 t^2} \\
& =\frac{\pi}{2}\left[\frac{\tan ^{-1}(\sqrt{3} t)}{\sqrt{3}}\right]_0^1 \\
& =\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}}
\end{aligned}
$$

Solutions · 1