Degrees of Definite Integral

Relevant wiki: Definite Integrals

Let $I = \large \int_0^\pi x^3\ln\sin x\, dx$

Using properties of Definite Integrals, $\displaystyle\int_0^a f(x)\, dx = \int_0^a f(a-x)\,dx$ $\begin{aligned}I &= \int_0^\pi \left(\pi-x\right)^3 \ln\sin x\, dx \\ &= \int_0^\pi\left(\pi^3-3\pi^2x+3\pi x^2-x^3\right)\ln\sin x\, dx\\ &=\int_0^\pi \pi^3\ln\sin x\, dx + \int_0^\pi -3\pi^2x\ln\sin x\, dx+\int_0^\pi 3\pi x^2\ln\sin x\, dx +\int_0^\pi -x^3\ln\sin x\, dx\\ &=I_1 + I_2 + I_3+I_4\end{aligned}$

Now, $\displaystyle I_1 = \pi^3 \int_0^\pi \ln\sin x\, dx$ , which is a standard integral $\int_0^\pi \ln\sin x\, dx = 2\int_0^{\pi/2} \ln\sin x\, dx = -\pi\ln 2$

So, $I_1 = -\pi^4 \ln 2$

Taking $I_2$ , using a formula I that I proved in my previous question , $\int_0^\pi x f(\sin x)\, dx = \dfrac\pi2 \int_0^{\pi} f(\sin x)\, dx\\\begin{aligned}\Rightarrow I_2 &= -3\pi^2 \int_0^\pi x\ln\sin x \, dx \\ &= -\dfrac {3\pi^3}2 \int_0^\pi \ln \sin x\, dx\\ &= \dfrac{3\pi^4}2\ln 2\end{aligned}$

For $I_3$ , we use $\ln ab = \ln a+\ln b$ , $\ln \sin x = \ln\left(\sqrt{2}\sin x\right) - \dfrac 12 \ln 2$ $\begin{aligned}I_3 &= 3\pi\int_0^\pi x^2\ln \sin x\, dx\\ &= 3\pi\int_0^\pi x^2\left(\ln\left(\sqrt{2}\sin x\right) - \dfrac 12 \ln 2\right)\, dx\\ &=3\pi k - \dfrac{3\pi}2\int_0^\pi \ln2\, dx\\ &=3\pi k-\dfrac{\pi^4}2 \ln 2\end{aligned}$

Now, simple $I_4 = -I$

Hence, $\begin{aligned}I& = -\pi^4 \ln 2+ \dfrac32\pi^4\ln 2+3\pi k-\dfrac12\pi^4\ln2 - I\\ \Rightarrow 2I &= \pi^4\ln 2\left(-1+\dfrac 32-\dfrac 12\right) + 3\pi k\\ \large I &= \large \boxed{\dfrac{3\pi}2 k}\end{aligned}$