Integrate $\frac{x}{\sin 2x}$ \begin{align}\int_{{\pi}\over5}^{{3\pi}\over10}\frac{x}{\sin2x}\,dx\end{align}
This integral came up while learning integration using Leibnitz rule. What has been tried is taking the integral as:
\begin{align}I(a)=\frac{1}{2}\int_{\pi\over5}^{3\pi\over10}\frac{\arctan (a \tan x)}{\sin x\cos x}\,dx\end{align} which gives $I'(a)$ as:
\begin{align}I'(a)=\frac{1}{2}\int_{2\pi\over5}^{3\pi\over5}\frac{dx}{\cos^2x(1+a^2\tan^2x)}=\frac{1}{2a}\arctan{\frac{a\tan \frac{3\pi}{10}-a\tan{\pi\over5}}{1+a^2\tan{\pi\over5}\cot{\pi\over5}}}\end{align}
\begin{align}=\frac{1}{2a}\arctan{\frac{a(\tan \frac{3\pi}{10}-\tan{\pi\over5})}{1+a^2}}\end{align}
Proceeding after this seems uncertain.
 A: A simpler approach using the following property of definite integral:

$$\int_{a}^ {b} f(x) dx=\int_{a} ^{b} f(a+b-x)dx$$

\begin{align}
I&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{x}{\sin{2x}}\:dx\\
&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{{\pi \over 5}+{3 \pi \over 10}-x}{\sin\left({2\left({\pi \over 5}+{3 \pi \over 10}-x\right)}\right)}\: dx\\
&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{\frac{\pi}{2}-x}{\sin{2\left(\frac{\pi}{2}-x\right)}}\:dx\\
&=\int_{\pi \over 5}^{3 \pi \over 10}\frac{\frac{\pi}{2}}{\sin{2x}}\:dx -\int_{\pi \over 5}^{3 \pi \over 10}\frac{x}{\sin{2x}}\:dx\\
&=\frac{\pi}{2}\int_{\pi \over 5}^{3 \pi \over 10}\frac{dx}{\sin{2x}} -I\\
\implies 2I &=\frac{\pi}{2}\int_{\pi \over 5}^{3 \pi \over 10} \csc{2x}\: dx\\
&=\frac{\pi}{2}\frac{1}{2} \left[\ln\left|\tan\left(\frac{2x}2\right)\right|\right]_{\pi \over 5}^{3 \pi \over 10}\\
&=\frac{\pi}{4} \left(\ln\left|\tan\left(\frac{3\pi}{10}\right)\right|-\ln\left|\tan\left(\frac{\pi}{5}\right)\right|\right)\\
\implies I&=\frac{\pi}{8} \ln\left|\frac{\tan\left(\frac{3\pi}{10}\right)}{\tan\left(\frac{\pi}{5}\right)}\right|\\
&\approx 0.250901
\end{align}
