Integrate without expansion?

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\\ % &= \sum_{i=0}^{10} \frac{(-1)^iC(10,i)}{2i+1}, \end{align}

where $$C(n,m) = \frac{n!}{m!(n-m)!}.$$

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:

$$\begin{align*} \int_0^1(1-x^2)^{10}\,\mathrm dx &= \frac{1}{2} \int_0^1 (1-u)^{10}u^{-1/2}\,\mathrm du,\qquad x=\sqrt{u} \\ &= \frac{1}{2} \int_0^1 (1-u)^{11-1}u^{1/2-1}\,\mathrm du\\ &= \frac{1}{2}\cdot\frac{\Gamma(11)\Gamma(1/2)}{\Gamma(11+1/2)}\\ &= \frac{1}{2}\cdot\frac{10!\sqrt{\pi}}{\frac{22!}{4^{11}\cdot11!}\sqrt{\pi}}\\ &= \frac{262144}{969969}. \end{align*}$$

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 $$\int_0^{\frac \pi 2}\cos^{n}(t)\,dt=\frac{\sqrt{\pi }}2 \frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma \left(\frac{n+2}{2}\right)}$$