Answer
$\displaystyle \frac{1}{15}$
Work Step by Step
$I=\displaystyle \int_{0}^{\pi/2}\cos 5t\cos 10tdt$
To evaluate the integral (c) $\displaystyle \int\cos mx\cos nxdx,$
$use $(c) $\displaystyle \quad \cos A\cos B=\frac{1}{2}[\cos(A-B)+\cos(A+B)]$
$\displaystyle \cos 10t\cos 5t=\frac{1}{2}[\cos(5t)+\cos(15t)]$
$I=\displaystyle \frac{1}{2}\int_{0}^{\pi/2}[\cos(5t)+\cos(15t)]dt$
$=\displaystyle \frac{1}{2}\left[\frac{1}{15}\sin(15t)+\frac{1}{5}\sin(5t)\right]_{0}^{\pi/2}$
$(\displaystyle \frac{15}{2}\pi=\frac{3\pi}{2}+6\pi, \quad \sin(\frac{15}{2}\pi)=-1$
$\displaystyle \frac{5}{2}\pi=\frac{\pi}{2}+2\pi, \quad\sin(\frac{5}{2}\pi)=+1)$
$=\displaystyle \frac{1}{2}[(-\frac{1}{15}+\frac{1}{5})-0]$
$=\displaystyle \frac{1}{2}\cdot\frac{2}{15}$
$=\displaystyle \frac{1}{15}$
