<s>[INST] 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? 
 [/INST] 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? 
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