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Let $I=\int_{1}^{\frac{\pi}{2}} \cos ^{2} x d x$

$\int \cos ^{2} x d x=\int\left(\frac{1+\cos 2 x}{2}\right) d x=\frac{x}{2}+\frac{\sin 2 x}{4}=\frac{1}{2}\left(x+\frac{\sin 2 x}{2}\right)=\mathrm{F}(x)$

By second fundamental theorem of calculus, we obtain

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

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

$=\frac{1}{2}\left[\frac{\pi}{2}+0-0-0\right]$

$=\frac{\pi}{4}$