**Problem**: Evaluate the improper integral $ I = \displaystyle \int_0^{\pi} \ln{(\sin{x})} dx $.

**Solution**: As the topic suggests, we will look for a symmetry to simplify the problem. Notice the identity $ \displaystyle \int_0^a f(x) dx = \int_0^a f(a-x) dx $ (since we're just integrating across the interval from different "directions." Using this, we have

$ \displaystyle I = \int_{0}^{\frac{\pi}{2}}\ln{(\sin{x})}dx = \int_{0}^{\frac{\pi}{2}}\ln{\left(\sin{\left(\frac{\pi}{2}-x\right)}\right)}dx = \int_{0}^{\frac{\pi}{2}}\ln{(\cos{x})}dx $.

Adding the old integral to the new one, we have

$ \displaystyle 2I = \int_{0}^{\frac{\pi}{2}}\ln{(\sin{x})}dx+\int_{0}^{\frac{\pi}{2}}\ln{(\cos{x})}dx = \int_{0}^{\frac{\pi}{2}}\ln{\left(\frac{1}{2}\sin{2x}\right)}dx $

from the property of logs and the double-angle identity (here it is again!). But in fact this expression is simply

$ \displaystyle 2I =-\frac{\pi}{2}\ln{2}+\frac{1}{2}\int_{0}^{\pi}\ln{(\sin{u})}du $ (*)

by the substitution $ u = 2x $. Taking into the account of the symmetry of $ \sin{x} $ from $ 0 $ to $ \pi $, we get $ \sin{x} = \sin{(\pi-x)} $ so

$ \displaystyle \frac{1}{2}\int_{0}^{\pi}\ln{(\sin{u})}du = \int_{0}^{\frac{\pi}{2}}\ln{(\sin{u})}du = I $.

Plugging back into (*) we obtain $ \displaystyle 2I =-\frac{\pi}{2}\ln{2}+I $ so we get $ I = -\frac{\pi}{2}\ln{2} $. QED.

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Comment: The function is not nice to actually integrate; it involves the polylogarithm function if you try here. This is an important example of how symmetry can help a ton in integration because there are so many functions that cannot be integrated with elementary functions but can be evaluated over an interval through different techniques.

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Practice Problem: Evaluate $ \displaystyle \int_0^{\frac{\pi}{2}} \sin^4{x} dx $ without actually finding the antiderivative.

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