Solve $\int_0^1((1-x)^8x^{11}-(1-x)^{11}x^8)dx$ I have got to solve this definite integral, though I have no idea in which direction to go. Another exercise I had is to solve something similar that looks like:
$$\int_0^1((1-x)^{11}x^2)dx$$
so for this integral I could substitute $t=1-x$ and I get a much easier version of the integral. The same method I couldn't implement on the given integral $\int_0^1((1-x)^8x^{11}-(1-x)^{11}x^8)dx$ as it just doesn't work.
The only thing I could notice is:
$$\int_0^1((1-x)^8x^{11}-(1-x)^{11}x^8)dx=\int_0^1((1-x)^8x^8(x^3-(1-x)^3))dx$$
yet I'm stuck at here and I have no clue how to solve the integral. (sure there is the simplest solution just to open up the polynomial - but it's naive and foolish...)
thanks
 A: 
NOTE:
We recognize that the integral of interest is simply $B(12,9)-B(9,12)$, where $$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$
is the Beta Function.  Then, exploiting the property of Beta Function, $B(x,y)=B(y,x)$, we immediately find that the result is $0$.  
I thought it would be instructive to present a way forward for those  unfamiliar with the Beta Function.  To that end, we proceed. 


The result of the integral of interest can be generalized as follows.  Let $I(x,y)$ be the integral given by 
$$I(x,y)=\int_0^1 t^x (1-t)^y\,dt$$
Now, enforcing the substitution $t \to 1-t$ we find that
$$\begin{align}
I(x,y)&=\int_0^1 t^x \,(1-t)^y\,dt\\\\
&=\int_1^0 (1-t)^x\,t^y\,(-1)\,dt\\\\
&=\int_0^1 (1-t)^x\,t^y\,dt\\\\
&=I(y,x)\\\\
\end{align}$$
Therefore, $I(x,y)=I(y,x)$ or $I(x,y)-I(y,x)=0$.
A: The symmetry of the expressions suggests shifting to center the integral around $0$. Substituting $u=x-1/2$, you have
$$\int_{-1/2}^{1/2}\left(\left(\frac12-u\right)^8\left(\frac12+u\right)^{11}-\left(\frac12-u\right)^{11}\left(\frac12+u\right)^8\right)du$$
The integral is of an odd function over $[-a,a]$, and so its value is $0$.
A: Since people are giving alternate approaches to the $t = 1 - x$ substitution, here's another one:
Integrate by parts in ${\displaystyle \int_0^1 (1 - x)^8x^{11}\,dx}$, integrating the $(1 - x)^8$ and differentiating the $x^{11}$. The endpoint terms are zero, so we get
$$\int_0^1 (1 - x)^8x^{11}\,dx = {11 \over 9} \int_0^1 (1 - x)^9x^{10}\,dx$$
Doing the same thing two more times yields
$$\int_0^1 (1 - x)^8x^{11}\,dx = {11 *10 * 9 \over 9 * 10 * 11} \int_0^1 (1 - x)^{11}x^8\,dx$$
This can be rearranged as 
$$\int_0^1 \big((1 - x)^8x^{11} - (1 - x)^{11}x^8\big)\,dx = 0$$
