I'm pleased to say that this is the first POTW that I've posted that had a lot of participants. I'd like to say thank you again for participating. :)
Anyways, this week's question was correctly answered by: anemone, BAdhi, Deveno, jacobi, magneto, MarkFL and Pranav. You can find Mark's solution below.
[sp]1.) Let:
$$I=\int_0^a f(x)\,dx$$
Now, use the substitution:
$$x=a-u\,\therefore\,dx=-du$$
and we may state:
$$I=-\int_a^0 f(a-u)\,du$$
Now, consider that the anti-derivative form of the FTOC allows us to state:
$$\int_a^b g(x)\,dx=G(b)-G(a)=-\left(G(a)-G(b) \right)=-\int_b^a g(x)\,dx$$
Note: $$\frac{d}{dx}\left(G(x) \right)=g(x)$$
Hence:
$$I=\int_0^a f(a-u)\,du$$
Exchanging the dummy variable of integration from $u$ to $x$, we obtain:
$$I=\int_0^a f(a-x)\,dx$$
Thus, we may conclude:
$$\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$$
2.) Let:
$$I=\int_0^{\frac{\pi}{2}}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}\,dx$$
Using the result obtained in part 1.) and the co-function identities for sine and cosine, we may also write:
$$I=\int_0^{\frac{\pi}{2}}\frac{\cos^n(x)}{\cos^n(x)+\sin^n(x)}\,dx=\int_0^{\frac{\pi}{2}}\frac{\cos^n(x)}{\sin^n(x)+\cos^n(x)}\,dx$$
Adding the two expressions, we find:
$$2I=\int_0^{\frac{\pi}{2}}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}+\frac{\cos^n(x)}{\sin^n(x)+\cos^n(x)}\,dx$$
Combining terms in the integrand and reducing, there results:
$$2I=\int_0^{\frac{\pi}{2}}\frac{\sin^n(x)+\cos^n(x)}{\sin^n(x)+\cos^n(x)}\,dx= \int_0^{\frac{\pi}{2}}\,dx$$
Applying the FTOC, we get:
$$2I=\left[x \right]_0^{\frac{\pi}{2}}=\frac{\pi}{2}-0=\frac{\pi}{2}$$
Dividing through by 2, we have:
$$I=\frac{\pi}{4}$$
Hence, we may conclude:
$$\int_0^{\frac{\pi}{2}}\frac{\sin^n(x)}{\sin^n(x)+ \cos^n(x)}\,dx=\frac{\pi}{4}$$[/sp]
