Evaluate the integral of $ sin(x) * cos(x) $ around the unit circle

To evaluate the integral of $ \sin(x) * \cos(x) $ around the unit circle, we can use the double-angle identity:

$ \sin(x) \cos(x) = \frac{1}{2} \sin(2x) $

Now, we need to integrate from $ 0 $ to $ 2\pi $:

$ \int_{0}^{2\pi} \sin(x) \cos(x) \, dx = \int_{0}^{2\pi} \frac{1}{2} \sin(2x) \, dx $

Let $ u = 2x $, hence $ du = 2 \, dx $ and $ dx = \frac{1}{2} du $:

$ \int_{0}^{2\pi} \frac{1}{2} \sin(2x) \, dx = \frac{1}{2} \int_{0}^{4\pi} \sin(u) \frac{1}{2} \, du $

Combining constants:

$ \frac{1}{4} \int_{0}^{4\pi} \sin(u) \, du $

The integral of $ \sin(u) $ over one period is zero, and here we have two periods:

$ \frac{1}{4} \left(0\right) = 0 $

The integral evaluates to $ 0 $.

Using the double-angle identity for $ sin(x) cos(x) $:

$ sin(x) cos(x) = frac{1}{2} sin(2x) $

Integrate over the unit circle:

$ int_{0}^{2pi} sin(x) cos(x) , dx = frac{1}{2} int_{0}^{2pi} sin(2x) , dx $

Substitute $ u = 2x $, $ du = 2dx $:

$ frac{1}{2} int_{0}^{2pi} sin(2x) , dx = frac{1}{2} cdot frac{1}{2} int_{0}^{4pi} sin(u) , du = frac{1}{4} int_{0}^{4pi} sin(u) , du $

Since the integral of $ sin(u) $ over $ 0 $ to $ 4pi $ is zero:

$ frac{1}{4} imes 0 = 0 $

Using the identity $ sin(x) cos(x) = frac{1}{2} sin(2x) $:

$ int_{0}^{2pi} sin(x) cos(x) , dx = frac{1}{2} int_{0}^{2pi} sin(2x) , dx $

Substitute $ u = 2x $:

$ frac{1}{4} int_{0}^{4pi} sin(u) , du = 0 $

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