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

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

• Is $\pi-2x$ squared or raised to the $6$th power? Your post is inconsistent. Commented Jul 28 at 5:41
• About the last integral, you can expand the square bracket and evaluate normally, using the reduction formula(if needed) Commented Jul 28 at 5:43
• What is the source of the integral? Commented Jul 28 at 5:51
• Closed form by mathematica $$\frac{\pi^{15}}{180180}$$ Commented Jul 28 at 5:54
• Why do not you expand everything from the beginning and create a large polynomial? Commented Jul 28 at 6:13

$$B(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$ 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 $$I=\frac{\pi^{15}}{180180}.$$
• Respected Mr Gérard, How have you received $-4B(8,7)$ and $4B(9,7)$? Explain, please. Commented Jul 28 at 14:09
• Because $(1-2x)^2=1-4x+4x^2$ implyi $$x^6(1-x)^6(1-2x)^2=x^6(1-x)^6-4x^7(1-x)^6+4x^8(1-x)^6.$$ Commented Jul 29 at 7:47
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*}