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


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


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

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



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