Question

E_n = int((2*x/Pi)^(2*n)*cos(x), x = -(1/2)*pi .. (1/2)*pi)

I'm supposed to use integration by parts to make a reduction formula for E_n but it seems to just make it more complicated. Can somebody help me figure this out?

Answer 1

So, to confirm,
\[ E_n = \int (2x/\pi)^{2n} \cos x \ dx \]
Are there limits on this integral?

Answer 2

the limits are -1/2 pi to 1/2 pi. Didn't realize that maple didn't copy paste into this well. Sorry about that.

Answer 3

Ok,
\[ E_n = \int_{-\pi/2}^{\pi/2} (2x/\pi)^{2n} \cos x \ dx \]
Now integrating by parts, this is equal to
\[ \left. (2x/\pi)^{2n} \sin x \right|_{-\pi/2}^{\pi/2} - 2n \int_{-\pi/2}^{\pi/2} (2/\pi)^{2n}x^{2n-1} \sin x \ dx \]
\[ = 1^{2n}.1 - 1^{2n}(-1) + Ditto \]
\[ = 2 + Ditto \]

- Now integrate by parts again, reducing the power of x
- Then rewrite the resulting integral so it is a (constant)*E_{n-1}

Answer 4

Wow, thank you. I'm an idiot. I left all kinds of nonsense in instead of simplifying and made it harder than it was.