Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3{(2x)}}{\ln{\left(\csc{x}\right)}} \mathop{dx}$ Challenge problem by friend is $$\int_0^{\frac{\pi}{2}} \frac{\sin^3{(2x)}}{\ln{\left(\csc{x}\right)}} \mathop{dx}$$
I know you can write $\ln{\left(\csc{x}\right)}=-\ln{\sin{x}}$ and $\sin{(2x)}=2\sin{(x)}\cos{(x)}$.  I tried rewriting the integral but then could not go further.  Even Wolfram Alpha (https://www.wolframalpha.com/input/?i=integral+of+sin%5E3%282x%29%2F%28log%28csc%28x%29%29%29+dx+from+0+to+pi%2F2) could not get a closed form!?  Is it even possible.
 A: You have the correct idea of expanding $\sin{(2x)}$:
$$I=8\int_0^{\frac{\pi}{2}} \frac{\sin^3{x}\cos^3{x}}{\ln{\left(\csc{x}\right)}} \; dx$$
Substituting $u=\ln{\left(\csc{x}\right)}$ will be very helpful:
$$I=8\int_{\infty}^0 \frac{\sin^3{x}\cos^3{x}}{u} \cdot \frac{du}{-\cot{x}}$$
$$=8\int_{0}^{\infty} \frac{\sin^4{x}\left(1-\sin^2{x}\right)}{u} \; du$$
$$=8\int_{0}^{\infty} \frac{e^{-4u}\left(1-e^{-2u}\right)}{u} \; du$$
$$=8\int_{0}^{\infty} \frac{e^{-4u}-e^{-6u}}{u} \; du$$
Now, this is a simple application of the Frullani integral:
$$8 \cdot \ln{\left(\frac{-6}{-4}\right)}=\boxed{8\ln{\left(\frac{3}{2}\right)}}$$
So, it turns out that this integral does have a closed form expression.
A: Take
$$I(a) = \int_0^{\pi /2} \frac{\sin^32x}{\ln \sin x}e^{-a\ln\sin x}\,dx$$
Differentiate w.r.t $a$ and use $\sin 2x = 2\sin x\cos x$ to get:
$$I'(a) = -8\int_0^{\pi /2} \sin^{3-a}x\cos^3 x\,dx$$
Put $\sin x = t$ to get:
$$I'(a) = -8\int_0^1t^{3-a}(1-t^2)\,dt$$
$$\implies I'(a) = 8\left(\frac1{a-4}-\frac{1}{a-6}\right)$$
Integrating this w.r.t $a$
$$\implies I(a) = 8\ln\left(\frac{a-4}{a-6}\right) + C$$
Now, as $a \rightarrow \infty$, $I \rightarrow 0$
$$\implies C = 0$$
$$\implies -I(0) = -8\ln\frac{2}{3} = \boxed{8\ln \left(\frac{3}{2}\right)}$$
