Question

Challenge of the day:

Give a closed-form expression for the following integral:
\[
\int_0^{\frac{\pi}{2}}\frac{\sin^n(x)}{\cos^n(x)+\sin^n(x)}\;dx
\]
For all \(n\in\mathbb{Z}\)

And, of course, prove it.

Nicest solution gets a medal and large props. I will post my own solution in 48 hours.

Answer 1

No attempts? :(

Answer 2

Last bump for this question, and I'll post up the answer.

Answer 3

I just saw it now, maybe if you had posted this before the weekend I would've attempted, but if it's too late, then it's too late. I don't have a pen with me and I'm at the library, so not going to give a response today that's for sure.

Answer 4

I can postpone it, if you wish, so long as there's an attempt.

I did post it two days ago, though.

@Kainui

Answer 5

Haha, well I want to try it, but I doubt I'll remember, I have a busy weekend ahead of me I'm afraid. Just post the answer and I'll be here to enjoy it at least haha.

Answer 6

That works. @Kainui

Answer 7

The most interesting part about this integral is that it's near-impossible to evaluate directly, but can be evaluated under inspection.

So, we generalize. Let:
\[
f(x)=\sin^n(x)
\]And
\[
g(x)=\cos^n(x)
\]And, let:
\[
D=\frac{\pi}{2}
\]Note that we have the property that, in general \(f(x)=g(D-x)\). So, we now have:
\[
\int_0^D\frac{f(x)}{g(x)+f(x)}\;dx\equiv S
\]Now, note the integral:
\[
\int_0^D \frac{g(x)}{f(x)+g(x)}\;dx
\]Under u-substitution, we have:
\[
\begin{align}
u&=D-x\\
du&=-dx
\end{align}
\]Hence our integral becomes:
\[
-\int_D^0\frac{g(u)}{f(u)+g(u)}\;du=\int_0^D\frac{g(u)}{f(u)+g(u)}\;du=\int_0^D\frac{f(x)}{g(x)+f(x)}\;dx=S
\]
So:
\[
\int_0^D\frac{f(x)}{g(x)+f(x)}\;dx+\int_0^D\frac{g(u)}{f(u)+g(u)}\;du=2S
\]Note that the choice of variable of integration is arbitrary, so we have:
\[
\int_0^D\frac{f(x)}{g(x)+f(x)}\;dx+\int_0^D\frac{g(u)}{f(u)+g(u)}\;du=\int_0^D\frac{f(x)+g(x)}{g(x)+f(x)}\;dx=
\]\[
\int_0^D1\;dx=D=2S
\]So:
\[
S\equiv\int_0^D\frac{f(x)}{g(x)+f(x)}\;dx=\frac{D}{2}
\]Which is fantastic, because it works for any case with the above properties. Now, we previously defined \(D=\frac{\pi}{2}\), giving us:
\[
\int_0^\frac{\pi}{2}\frac{\sin^n(x)}{\cos^n(x)+\sin^n(x)}\;dx=\frac{\pi}{4}, \;\forall n\in \mathbb{Z}
\]

Answer 8

There we go. Sorry, the site went down a moment, for me.