Compute the integral

Compute the following integral:

$\int_{- \pi}^x \left( \frac{1}{2} + \cos t\right)^2 \, dt.$

We have

$\begin{align*} \int_{-\pi}^x \left( \frac{1}{2} + \cos t \right)^2 \, dt &= \int_{-\pi}^x \left( \frac{1}{4} + \cos t + \cos^2 t \right) \, dt \\ &= \frac{1}{4} \int_{-\pi}^x \, dt + \int_{-\pi}^x \cos t \, dt + \int_{-\pi}^x \cos^2 t \, dt \\ &= \frac{1}{4} (x + \pi) + (\sin x - \sin (-\pi)) + \int_{-\pi}^x \frac{1}{2}(1 + \cos (2t)) \, dt \\ &= \frac{1}{4} (x + \pi) + \sin x + \frac{1}{2} (x + \pi) + \frac{1}{4} \int_{-2 \pi}^{2x} \cos t \, dt \\ &= \frac{3}{4} (x+ \pi) + \sin x + \left(\frac{1}{4} (\sin x) \biggr \rvert_{-2 \pi}^{2x}\right) \\ &= \frac{3}{4} (x + \pi) + \sin x + \frac{1}{4} \sin (2x). \end{align*}$

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