Is integration by parts the best method for $\int_0^1 x^3(1-x)^6 dx$? It came up when finding a constant such that the integral is equal to 1 and thus behaves like a pdf. I used the parts method but have made an error, just curious how others might approach the problem.
 A: By symmetry, 
\begin{align}
\int_0^1 x^3(1-x)^6\,\mathrm{d}x &= -\int_1^0x^6(1-x)^3\mathrm{d}x\\
&=-\int_0^1 (x-1)^3x^6\,\mathrm{d}x \\
&=-\int_0^1 x^9-3x^8+3x^7-x^6 \,\mathrm{d}x\\
&=-\left(\frac{1}{10}-\frac{1}{3}+\frac{3}{8}-\frac{1}{7}\right)+0\\\\
&=-\frac{84-280+315-120}{840}\\
&=\boxed{\displaystyle\frac{1}{840}}
\end{align}
A: Along the lines of Kaj, I think the fastest is to just expand:
$$
\int x^3(1-x)^6 = \int (1-u)^3u^6=\int u^6-3u^7+3u^8-u^9=\frac{1}{7}-\frac{3}{8}+\frac{3}{9}-\frac{1}{10}.
$$
A: If we realize the integral is part of the Beta Distribution:
$$\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1},0<x<1$$
We find $\alpha=4,\beta=7$. We also know the integral of any pdf equals $1$ and thus this our coefficient must be the inverse of the result.
A: The beta function is the best idea.
$$\beta(a, b) = \int_{0}^{1} x^{a-1}(1-x)^{b-1} dx$$
For here $I$, let $a = 4, b = 7$
Using:
$$\beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
$$\beta(4, 7) = \frac{\Gamma(4)\Gamma(7)}{\Gamma(11)} = \frac{3!6!}{10!}$$
$$= \frac{3\cdot2\cdot1}{10\cdot9\cdot8\cdot7} = \frac{1}{840}$$
A: This way might be faster, but it ultimately depends on your personal preference.
Let $u = 1-x$.  Then the integral becomes:
$$\int (u-1)^3u^6 \  \text{d}u$$
And that expansion is much easier to work with than the original!
A: using that $$x^3(1-x)^6=x^9-6 x^8+15 x^7-20 x^6+15 x^5-6 x^4+x^3$$ we can calculate the integral without the beta function, we get
$$\int_{0}^{1} x^9-6 x^8+15 x^7-20 x^6+15 x^5-6 x^4+x^3dx=\frac{1}{840}$$
