Let $f(x)$ be a continuous function

[Chris L T521]

Thanks again to those who participated in last week's POTW! Here's the first University POTW of 2014!

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Problem: Let $f(x)$ be a continuous function.

1. Show that $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.
2. Use (1) to show that \[\int_0^{\pi/2}\frac{\sin^n x}{\sin^n x+\cos^n x}\,dx = \frac{\pi}{4}\] for all positive numbers $n$.

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[Chris L T521]

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:

\(\displaystyle I=\int_0^a f(x)\,dx\)

Now, use the substitution:

\(\displaystyle x=a-u\,\therefore\,dx=-du\)

and we may state:

\(\displaystyle I=-\int_a^0 f(a-u)\,du\)

Now, consider that the anti-derivative form of the FTOC allows us to state:

\(\displaystyle \int_a^b g(x)\,dx=G(b)-G(a)=-\left(G(a)-G(b) \right)=-\int_b^a g(x)\,dx\)

Note: \(\displaystyle \frac{d}{dx}\left(G(x) \right)=g(x)\)

Hence:

\(\displaystyle I=\int_0^a f(a-u)\,du\)

Exchanging the dummy variable of integration from $u$ to $x$, we obtain:

\(\displaystyle I=\int_0^a f(a-x)\,dx\)

Thus, we may conclude:

\(\displaystyle \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\)

2.) Let:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}\,dx\)

Using the result obtained in part 1.) and the co-function identities for sine and cosine, we may also write:

\(\displaystyle I=\int_0^{\frac{\pi}{2}}\frac{\cos^n(x)}{\cos^n(x)+\sin^n(x)}\,dx=\int_0^{\frac{\pi}{2}}\frac{\cos^n(x)}{\sin^n(x)+\cos^n(x)}\,dx\)

Adding the two expressions, we find:

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}+\frac{\cos^n(x)}{\sin^n(x)+\cos^n(x)}\,dx\)

Combining terms in the integrand and reducing, there results:

\(\displaystyle 2I=\int_0^{\frac{\pi}{2}}\frac{\sin^n(x)+\cos^n(x)}{\sin^n(x)+\cos^n(x)}\,dx= \int_0^{\frac{\pi}{2}}\,dx\)

Applying the FTOC, we get:

\(\displaystyle 2I=\left[x \right]_0^{\frac{\pi}{2}}=\frac{\pi}{2}-0=\frac{\pi}{2}\)

Dividing through by 2, we have:

\(\displaystyle I=\frac{\pi}{4}\)

Hence, we may conclude:

\(\displaystyle \int_0^{\frac{\pi}{2}}\frac{\sin^n(x)}{\sin^n(x)+ \cos^n(x)}\,dx=\frac{\pi}{4}\)[/sp]