Question:
$\int_{0}^{\frac{\pi}{2}} \cos 2 x d x$
Solution:
Let $I=\int_{0}^{\frac{\pi}{2}} \cos 2 x d x$
$\int \cos 2 x d x=\left(\frac{\sin 2 x}{2}\right)=\mathrm{F}(x)$
By second fundamental theorem of calculus, we obtain
$\begin{aligned} I &=\mathrm{F}\left(\frac{\pi}{2}\right)-\mathrm{F}(0) \\ &=\frac{1}{2}\left[\sin 2\left(\frac{\pi}{2}\right)-\sin 0\right] \\ &=\frac{1}{2}[\sin \pi-\sin 0] \\ &=\frac{1}{2}[0-0]=0 \end{aligned}$
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