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