Evaluation of $\int^\pi_{0}(\pi-x)^6x^6(\pi-2x)^2dx$

Question

Evaluation of

$\displaystyle \int^\pi_{0}(\pi-x)^6x^6(\pi-2x)^2dx$

What I try :

Let $\displaystyle I =\int^{\pi}_{0}(\pi-x)^6x^6(\pi-2x)^2dx$

Using property

$\displaystyle \int^{a}_{0}f(x)dx=\int^a_0f(a-x)dx$

$\displaystyle I =\int^{\pi}_{0}x^6(\pi-x)^6(\pi-2x)^2dx$

$\displaystyle I =2\int^{\frac{\pi}{2}}_{0}x^6(\pi-x)^2(\pi-2x)^2dx$

When I try using Trigonometric Substution ,

$\displaystyle x =\pi\sin^2\theta$ and $\displaystyle dx=2\pi\sin \theta\cos \theta d\theta$

$\displaystyle I =4\int^{1}_{0}\pi^6\sin^{12}\theta\cdot \pi^6\sin^{12}\theta(\pi-2\pi\sin^2\theta)^2\sin \theta\cos\theta d\theta$

Now I did not understand how can I solve further, Help me

Answer 1

$B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$ implies $$\int_0^1(1-x)^6x^6(1-2x)^2dx=B(7,7)-4B(8,7)+4B(9,7)=\frac{6!6! 14}{15!}=\frac{1}{180180}.$$ The change of variable $x=\pi t$ in the initial integral $I$ gives $I=\frac{\pi^{15}}{180180}.$

Answer 2

Rather than reflecting about $x=\dfrac\pi2$, consider a translation then a power reduction to reveal a beta integral.

$$\begin{align*} & \int_0^\pi (\pi-x)^6 x^6 (\pi-2x)^6 \, dx \\ &= \int_{-\tfrac\pi2}^\tfrac\pi2 \left(y-\frac\pi2\right)^6 \left(y+\frac\pi2\right)^6 (2y)^6 \, dy & x=y+\frac\pi2 \\ &= 2^6 \int_{-\tfrac\pi2}^\tfrac\pi2 y^6 \left(y^2-\frac{\pi^2}4\right)^6 \, dy \\ &= 2^{6+1} \int_0^\tfrac\pi2 y^6 \left(y^2-\frac{\pi^2}4\right)^6 \, dy \\ &= \frac{\pi^{19}}{2^{13}} \int_0^1 z^{5/2} (1-z)^6 \, dz & y=\frac\pi2 \sqrt z \\ &= \frac{\pi^{19}}{2^{13}} \operatorname{B}\left(\frac72,7\right) \end{align*}$$