Answer
$\frac{16-5\pi^{2}}{8}$
Work Step by Step
Let $I=\displaystyle\int_{0}^{\frac{\pi}{2}}(2\cos(x)-5x)dx$.
We have:
$I=\displaystyle\int_{0}^{\frac{\pi}{2}}2\cos(x)dx+\int_{0}^{\frac{\pi}{2}}(-5x)dx$
$=2\displaystyle\int_{0}^{\frac{\pi}{2}}\cos xdx-5\int_{0}^{\frac{\pi}{2}}x dx$
Use $\int_0^{\frac{\pi}{2}}\cos x=1$:
$I=2(1)-5\displaystyle\int_{0}^{\frac{\pi}{2}}xdx$
Use that $\int_a^b xdx=\frac{b^2-a^2}{2}$:
$I=2-5\frac{\left(\frac{\pi}{2}\right)^{2}-0^{2}}{2}=\frac{16-5\pi^{2}}{8}$
