## Tuesday, May 16, 2006

### See What The Integrator Has To Say. Topic: Calculus/Trigonometry. Level: Olympiad.

Problem: Evaluate $\displaystyle \int_0^\pi \frac{x\sin{x}}{1+\cos^2{x}} dx$.

Solution: Now this does not have a straightforward integrate it solution, as The Integrator will show you. So we think... symmetry.

Note that $\sin{x} = \sin{(\pi-x)}$ and $\cos^2{x} = \cos^2{(\pi-x)}$. So that means

$\displaystyle \int_0^\pi \frac{x\sin{x}}{1+\cos^2{x}} dx = \frac{1}{2} \int_0^\pi \frac{(x+(\pi-x)) \sin{x}}{1+\cos^2{x}} dx = \frac{\pi}{2} \int_0^\pi \frac{\sin{x}}{1+\cos^2{x}} dx$.

But this integral is much more friendly, and we can easily make the substitution $u = \cos{x}$ so $du = -\sin{x} dx$. Our integral becomes

$\displaystyle -\frac{\pi}{2} \int_1^{-1} \frac{du}{1+u^2} = -\frac{\pi}{2} [\arctan{u}]_1^{-1} = -\frac{\pi}{2} \left(-\frac{\pi}{4}-\frac{\pi}{4}\right) = \frac{\pi^2}{4}$.

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Comment: Here we see the power of symmetry in integration - there's no clear way to evaluate the given integral so we try to make it into something we know. Using symmetry works well when functions are symmetric around the origin or you have inverse functions.

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Practice Problem: Evaluate $\displaystyle \int_0^1 \left(\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}\right) dx$.