Hi, I was doing a question with this simple integral

$\displaystyle 3a\int^{\frac{\pi}{2}}_0cosxsinxdx$

so rather than do it by parts (which probably would turn out to be double back on itself so you get the integral again), I opted for using sin2x = 2sinxcosx

so that means the integral simplified to

$\displaystyle 3a\int^{\frac{\pi}{2}}_0\frac{1}{2}sin2x$

which integrates to

$\displaystyle 3a[-\frac{1}{4}cos2x]^{\frac{\pi}{2}}_0$

so wack in the limits and I get 3a[1-1] = 0

but that isn't right is it?

The answer scheme suggest $\displaystyle 3a\int^{\frac{\pi}{2}}_0cosxsinxdx$ = $\displaystyle 3a[\frac{1}{2}sin^{2}2x]^{\frac{\pi}{2}}_0$

but i) How come my method doesn't work

ii) How can the mark scheme do that. You can't just do that!?

Thanks