Where k is a positive integer, evaluate:
$\displaystyle \int_{-1}^{1} \! (1 - x^2)^k \, dx.$
Possibly make a change of variables? Any help?
~Thanks
It can't be zero ever, because $\displaystyle (1-x^2)^k$ is an even function on a symmetric interval, so $\displaystyle \int_{-1}^1(1-x^2)^k\,dx=2\int_0^1(1-x^2)^k\,dx$. Since $\displaystyle 0\leq1-x^2\leq1$ on $\displaystyle [0,1]$, this integral is always strictly positive.
Furthermore, the antiderivative is not $\displaystyle \frac{1}{2k+2}(1-x^2)^{k+1}$. The derivative of that function is $\displaystyle -x(1-x^2)^k$.
Use the substitution $\displaystyle x = \sin{\theta}$ so that $\displaystyle dx = \cos{\theta}\,d\theta$.
Also note that $\displaystyle 1 - x^2 = 1 - \sin^2{\theta} = \cos^2{\theta}$.
So $\displaystyle \int{(1 - x^2)^k\,dx} = \int{\cos^{2k}{\theta}\,\cos{\theta}\,d\theta}$
$\displaystyle = \int{\cos^{2k + 1}{\theta}\,d\theta}$.