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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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