Integrate without expansion? I want to evaluate 
$$ \int_0^1 ( 1 - x^2)^{10} dx $$
One way I can do this is by expanding out $(1 - x^2)^{10}$ term by term, but is there a better way to do this? 
 A: Similar to dxdydz's answer, let $x=\sin(t)$
$$\int_0^1 ( 1 - x^2)^{10}\, dx=\int_0^{\frac \pi 2}\cos^{21}(t)\,dt$$ and remember that
$$\int_0^{\frac \pi 2}\cos^{n}(t)\,dt=\frac{\sqrt{\pi }}2 \frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma
   \left(\frac{n+2}{2}\right)}$$
A: When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $\int(k^2 - x^2)^n dx $. 
Using integration by parts:
$u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^{n-1}(-2x)dx $ and $v =x$
Thus, 
$$\int(k^2 - x^2)^n dx = x(k^2 - x^2)^{n-1} + 2n \int x^2(k^2 - x^2)^{n-1}  \\
= x(k^2 - x^2)^{n-1} + 2n \int \bigg[ k^2(k^2 - x^2)^{n-1} - (k^2 - x^2)^{n} \bigg] dx \\
=  x(k^2 - x^2)^{n-1} + 2n \int k^2(k^2 - x^2)^{n-1} dx - 2n \int (k^2 - x^2)^{n} dx
$$ 
From here you can see that:
$$2n\int(k^2 - x^2)^n dx +  \int(k^2 - x^2)^n dx  = x(k^2 - x^2)^{n-1} + 2n \int k^2(k^2 - x^2)^{n-1} dx - 2n \int (k^2 - x^2)^{n} dx $$
Thus,
$$\int(k^2 - x^2)^n dx = \frac{x(k^2 - x^2)^{n-1}}{2n+1} + \frac{2nk^2}{2n+1} \int (k^2 - x^2)^{n-1} dx$$
For your problem, $k=1$ and $n=10$. Can you take it from here? 
A: This answer is exactly what the asker is NOT asking for. However, I believe it will be useful to show the asker that expanding out "term-by-term" is not actually that hard. 
We have 
\begin{align}
\int_0^1 (1-x^2)^{10}dx &= \int_0^1 \sum_{i=0}^{10} C(10,i)(-x^2)^idx \\
%
& = \sum_{i=0}^{10} (-1)^iC(10,i)\int_0^1 x^{2i}dx\\
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&= \sum_{i=0}^{10} \frac{(-1)^iC(10,i)}{2i+1},
\end{align}
where $$C(n,m) = \frac{n!}{m!(n-m)!}.$$
A: An alternate approach is through the beta function. We start by u-substitution and try to make our way towards an integral form of the beta function:
$$\begin{align*}
\int_0^1(1-x^2)^{10}\,\mathrm dx &= \frac{1}{2} \int_0^1 (1-u)^{10}u^{-1/2}\,\mathrm du,\qquad x=\sqrt{u} \\ 
 &= \frac{1}{2} \int_0^1 (1-u)^{11-1}u^{1/2-1}\,\mathrm du\\ 
 &= \frac{1}{2}\cdot\frac{\Gamma(11)\Gamma(1/2)}{\Gamma(11+1/2)}\\ 
 &= \frac{1}{2}\cdot\frac{10!\sqrt{\pi}}{\frac{22!}{4^{11}\cdot11!}\sqrt{\pi}}\\ 
 &= \frac{262144}{969969}.
\end{align*}$$
