Show that $$\int_0^{\pi/2}\frac x {\sin(x)+6x} \, dx =\frac{\pi}{2\sqrt2}\ln \left( \cot \frac \pi 8 \right)$$
I tried to apply a related theory for the integral using substitution for $x,$ since there is 0 as the lower interval, $x=\frac\pi2-x $
$$\int_0^{\pi/2}\frac{\frac\pi2-x}{\sin(\frac\pi2-x)+6(\frac\pi2-x)} \, dx$$
But what to do next? Is there a specific substitution when there is a denominator with trigonometric and algebraic function together?
Since $\frac{2}{\pi}x\leq \sin(x)\leq x$ for any $x\in\left[0,\frac{\pi}{2}\right]$ it follows that $$ \int_{0}^{\pi/2}\frac{x}{\sin(x)+6x}\,dx \in \left[\frac{\pi}{14},\frac{\pi^2}{12\pi+4}\right]$$ while $\frac{\pi}{2\sqrt{2}}\log\cot\frac{\pi}{8}$ is approximately $1$.
There must have been a typo since the RHS is approximately four times larger than the LHS.