
Evaluate
Evaluate $\displaystyle \int^{\frac{\pi}{2}}_{0}2cosx+3sin2xdx + \int^{\frac{5\pi}{6}}_{\frac{\pi}{2}}2cosx+3sin2x dx$
The ans to this question is $\displaystyle 4.25units^2$ which is different from my answer of $\displaystyle 8.25units^2$
What I did
$\displaystyle \int^{\frac{\pi}{2}}_{0}2cosx+3sin2xdx + \int^{\frac{5\pi}{6}}_{\frac{\pi}{2}}2cosx+3sin2x dx$ $\displaystyle =$ $\displaystyle [2sinx\frac{3}{2}cos2x]^{\frac{\pi}{2}}_{0}+[2sinx\frac{3}{2}cos2x]^{\frac{5\pi}{6}}_{\frac{\pi}{2}}$
$\displaystyle =(3.5(1.5))+(0.253.5)$
$\displaystyle =8.25units^2$


Maybe you got it right and the answer of 4.25 is wrong.
I got the answer of 8.25 as well.