Trigonometric reduction

[JamesGregory]

1. The problem statement, all variables and given/known data

Prove the following trigonometric reduction using integration by parts:


\int \sin^n x dx = - \frac{\ \sin^{n-1} x \cos x}{n} + \frac{\ n-1}{n} \int \sin^{n-2} x dx


2. The attempt at a solution

I tried using integration by parts by breaking up sin^n x into sin^(n-2) x sin^2 x and sin^(n-1) x sin x but couldn't seem to get either to work. If someone could just provide a useful link as opposed to typing the whole proof, that would be appreciated.

[HallsofIvy]

Did you consider induction on n?
It's obvious that this is true when n= 1.

Assume the statement is true for sink x and use integration by parts on
[tex]\int sin^{k+1} xdx= \int sin x sin^k x dx[/itex]
with u= cos x and dv= sink dx.

[JamesGregory]

This is high school ap calculus ab, and I don't think we've gotten to induction, but I do think I see what you're getting at. but shouldn't u and dv be parts of the original integrand or am I missing something obvious?

[Gib Z]

Ahh yes I think Halls just typoed, but I he should mean u = sin x, and dv = sin^k x.

[dextercioby]

In my hs, we got induction method 2 yrs before integral calculus...

[HallsofIvy]

This is a high school class, you haven't done proof by induction, but you have done integration? Am I the only one who thinks that is backwards?

[Gib Z]

I was about to say that in my post as well, but seeing as Australia does about everything backwards maybe I didn't know what i was talking about. Though mathwonk always like to say how some of his Calculus students don't know basic algebra and how it appalls him. Something's seriously wrong with that education system.