# Problem of the Week #93 - January 6th, 2014

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#### Chris L T521

##### Well-known member
Staff member
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

##### Well-known member
Staff member
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.

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:

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

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