Answer
\[\frac{1}{15}\]
Work Step by Step
Considering indefinite integral
Let
\[I=\int \cos 5t\:\cos 10t\:dt\]
\[I=\frac{1}{2}\int 2 \cos 5t\:\cos 10t\:dt\]
\[\left[2\cos A\cos B=\cos (A+B)+\cos (A-B)\right]\]
\[I=\frac{1}{2}\int \left[ \cos 15t\:+\cos (-5t)\right]\:dt\]
\[\left[\cos(-x)=\cos x\right]\]
\[I=\frac{1}{2}\int \left[ \cos 15t\:+\cos 5t\right]\:dt\]
\[I=\frac{1}{2}\left[\frac{1}{15} \sin 15t\:+\frac{1}{5}\sin 5t\right]\]
$I=\frac{1}{30}\sin 15t\:+\frac{1}{10}\sin 5t$ ___(1)
Let
\[I_1=\int_{0}^{\frac{π}{2}} \cos 5t\:\cos 10t\:dt=\left[\frac{1}{30}\sin 15t\:+\frac{1}{10}\sin 5t\right]_{0}^{\frac{π}{2}}\]
\[I_1=\frac{1}{30}\sin \frac{15π}{2}\:+\frac{1}{10}\sin \frac{5π}{2}-\frac{1}{30}\sin (0)\:-\frac{1}{10}\sin (0)\]
\[I_1=\frac{1}{30}\sin\left(8π- \frac{π}{2}\right)\:+\frac{1}{10}\sin \left(2π+\frac{π}{2}\right)\]
\[I_1=-\frac{1}{30}+\frac{1}{10}=\frac{-1+3}{30}\]
\[I_1=\frac{1}{15}\]
Hence \[I_1=\frac{1}{15}\].
