Work out $\int_{0}^{1}t^2\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]dt$ I would like to evaluate this

$$I=\int_{0}^{1}t^2\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]\mathrm dt$$

Using trig:
$$I=\int_{0}^{1}t^2\cos(2t\pi)\cdot \frac{1-\cos(2t\pi)}{\sin(2t\pi)}\cdot\ln[\sin(t\pi)]\mathrm dt$$
$$I=\int_{0}^{1}t^2\cdot \frac{\cos(2t\pi)-1+\sin^2(2t\pi)}{\sin(2t\pi)}\cdot\ln[\sin(t\pi)]\mathrm dt$$
$$I=\int_{0}^{1}t^2\cot(2t\pi)\ln[\sin(t\pi)]\mathrm dt-\int_{0}^{1}t^2\csc(2t\pi)\ln[\sin(t\pi)]\mathrm dt+\int_{0}^{1}t^2\sin(2t\pi)\ln[\sin(t\pi)]\mathrm dt$$

$$\int_{0}^{1}t^2\cot(2t\pi)\ln[\sin(t\pi)]\mathrm dt=\sum_{n=0}^{\infty}\frac{(-1)^nB_{2n}}{(2n)!}(2\pi)^{2n-1}\int_{0}^{1}t^{2n+1}\ln[\sin(t\pi)]\mathrm dt$$
$$\int_{0}^{1}t^2\sin(2t\pi)\ln[\sin(t\pi)]\mathrm dt=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(2\pi)^{2n+1}\int_{0}^{1}t^{2n+3}\ln[\sin(t\pi)]\mathrm dt$$

This integral it is too difficult to reduce
$$J=\int_{0}^{1}t^{2n+3}\ln[\sin(t\pi)]\mathrm dt$$
I can't continue
 A: $$\color{blue}{I = \frac{{\ln 2}}{2\pi}(1 - \ln 2)} $$

We have
$$\begin{aligned}
I &= \frac{1}{{{\pi ^3}}}\int_0^\pi {{x^2}\cos (2x)\tan x\ln (\sin x)dx} \\
&= \frac{1}{{{\pi ^3}}}\int_{ - \pi /2}^{\pi /2} {{{\left( {\frac{\pi }{2} + x} \right)}^2}\cos (2x)\cot x\ln (\cos x)dx} \\
&= \frac{1}{{{\pi ^2}}}\int_{ - \pi /2}^{\pi /2} {x\cos (2x)\cot x\ln (\cos x)dx} 
\end{aligned}$$
Now invoke the formula:

When $a-b-c>0$, we have $$\int_{ - \pi /2}^{\pi /2} {{e^{iax}}{{( - 2i\sin x)}^b}{{(2\cos x)}^c}dx} = \sin \left( {\frac{\pi }{2}(a - b - c)} \right)\int_0^1 {{x^{(a - b - c)/2 - 1}}{{(1 - x)}^c}{{(1 + x)}^b}dx} $$
  For a proof, see here.

Differentiate both sides with respect to $a$ and $c$, then set $a=2,b=-1,c=1$ gives
$$i\int_{ - \pi /2}^{\pi /2} {\frac{{x{e^{2ix}}}}{{ - 2i\sin x}}(2\cos x)\ln (2\cos x)dx} =  \frac{\pi }{2}\int_0^1 {\left[ {\frac{{(1 - x)\ln x}}{{1 + x}} - \frac{{(1 - x)\ln (1 - x)}}{{1 + x}}} \right]dx} $$
You should have no difficulty in evaluating RHS, giving
$$\int_{ - \pi /2}^{\pi /2} {x\cot x\cos 2x\ln (2\cos x)dx} = \frac{{\pi {{\ln }^2}2}}{2}$$
I left you as an exercise to show that
$$\int_{ - \pi /2}^{\pi /2} x\cot x \cos 2x dx = \pi (\ln 2 - \frac{1}{2})$$
This completes the proof.
