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Hence, $f$ is even, and we need to only consider.

$$

f(\theta) = \frac{a_0}{2} + \sum_{n = 1}^{\infty}a_n\cos n\theta.

$$

$$

a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}|\sin\theta|d\theta = \frac{1}{\pi}\int_{0}^{\pi}\sin\theta d\theta - \frac{1}{\pi}\int_{-\pi}^0\sin\theta d\theta = \frac{2}{\pi}\int_{0}^{\pi}\sin\theta d\theta

$$

but when I solve this integral, I get 0.

How do I solve this integral?

$$

a_n =\frac{1}{\pi}\int_{-\pi}^{\pi}|\sin\theta|\cos n\theta d\theta = \frac{2}{\pi}\int_{-\pi}^{\pi}\sin\theta\cos n\theta d\theta

$$