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Question:

$\int_{0}^{\frac{\pi}{4}} \sin 2 x d x$

Solution:

Let $I=$ $\int_{0}^{\frac{\pi}{4}} \sin 2 x d x$

$\int \sin 2 x d x=\left(\frac{-\cos 2 x}{2}\right)=\mathrm{F}(x)$

By second fundamental theorem of calculus, we obtain

$I=\mathrm{F}\left(\frac{\pi}{4}\right)-\mathrm{F}(0)$

$=-\frac{1 \pi}{2}\left[\cos 2\left(\frac{}{4}\right)-\cos 0\right]$

$=-\frac{1 \pi}{2}\left[\cos \left(-\frac{1}{2}\right)-\cos 0\right]$

$=-\frac{1}{2}[0-1]$

$=\frac{1}{2}$

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