Mathematics - Area Under Curves Question with Solution | TestHub

At the intersection, $2\cos x=3\tan x$, so $2{\sin }^{2}x+3\sin x-2=0$ and $\sin x=1/2$, giving $x=\pi /6$. The area is ${\int }_0^{\pi /6}(2\cos x-3\tan x)dx=[2\sin x+3\ln (\cos x){]}_0^{\pi /6}=1+\frac{3}{2}\ln 3-3\ln 2$.