1. ## Evaluate the integral

Where k is a positive integer, evaluate:

$\int_{-1}^{1} \! (1 - x^2)^k \, dx.$

Possibly make a change of variables? Any help?

~Thanks

2. Originally Posted by Anonymous1
Where k is a positive integer, evaluate:

$\int_{-1}^{1} \! (1 - x^2)^k \, dx.$

Possibly make a change of variables? Any help?

~Thanks
You can expand it using the Binomial Theorem...

4. Originally Posted by chengbin
Look closer, the answer is 0 no matter what value of k.

$=\bigg [ \frac{1}{2k+2}(1-x^2)^{k+1} \bigg ]_{-1}^1$

$=\bigg ( \frac{1}{2k+2} (1-1)^{k+1} \bigg ) - \bigg ( \frac{1}{2k+2} (1-1)^{k+1} \bigg )$

$=0$
It can't be zero ever, because $(1-x^2)^k$ is an even function on a symmetric interval, so $\int_{-1}^1(1-x^2)^k\,dx=2\int_0^1(1-x^2)^k\,dx$. Since $0\leq1-x^2\leq1$ on $[0,1]$, this integral is always strictly positive.

Furthermore, the antiderivative is not $\frac{1}{2k+2}(1-x^2)^{k+1}$. The derivative of that function is $-x(1-x^2)^k$.

5. Originally Posted by redsoxfan325
It can't be zero ever, because $(1-x^2)^k$ is an even function on a symmetric interval, so $\int_{-1}^1(1-x^2)^k\,dx=2\int_0^1(1-x^2)^k\,dx$. Since $0\leq1-x^2\leq1$ on $[0,1]$, this integral is always strictly positive.

Furthermore, the antiderivative is not $\frac{1}{2k+2}(1-x^2)^{k+1}$. The derivative of that function is $-x(1-x^2)^k$.
I realized that after I posted.

6. I realized that you realized that after I posted.

7. Originally Posted by Prove It
You can expand it using the Binomial Theorem...
Really? The whole $x^2$ thing isn't an issue?

8. Nope. The binomial theorem works with anything.

9. Originally Posted by Anonymous1
Where k is a positive integer, evaluate:

$\int_{-1}^{1} \! (1 - x^2)^k \, dx.$

Possibly make a change of variables? Any help?

~Thanks
Use the substitution $x = \sin{\theta}$ so that $dx = \cos{\theta}\,d\theta$.

Also note that $1 - x^2 = 1 - \sin^2{\theta} = \cos^2{\theta}$.

So $\int{(1 - x^2)^k\,dx} = \int{\cos^{2k}{\theta}\,\cos{\theta}\,d\theta}$

$= \int{\cos^{2k + 1}{\theta}\,d\theta}$.