Question

integral from 0 to 4 pi of (5 sin x)+(2 cosx) dx

Answer 1

\[\int\limits_{0}^{4\Pi}(5sinx+2cosx)dx\]

Answer 2

I keep getting -20pi, what am I doing wrong?

Answer 3

Hmmm you shouldn't be ending up with a pi anywhere in your answer D:
Let's see what's going on...\[\Large \int\limits_0^{4\pi}5\sin x+2\cos x\;dx\]
Integrating gives us:\[\Large -5\cos x+2\sin x|_0^{4\pi}\]Sine of 4pi and sine of 0 give us 0 so we'll ignore that part.\[\Large -5\cos x|_0^{4\pi}\quad=\quad -5\left[\cos 4\pi-\cos 0\right]\quad=\quad ?\]