Answer:
5/2
Step-by-step explanation:
Let u = sin(t). Then this is the integral ...
[tex]\displaystyle\int_0^{\frac{\pi}{2}}{5u}\,du=\left.\dfrac{5u^2}{2}\right|_0^{\frac{\pi}{2}}=\dfrac{5}{2}(\sin(\frac{\pi}{2})^2-\sin(0)^2)=\dfrac{5}{2}(1-0)=\boxed{\dfrac{5}{2}}[/tex]
