Question

Integrate...

Answer 1

\[\LARGE \int\limits_{0}^{\frac{32 \pi}{3}}\sqrt{1+\cos2x}~dx\]

Answer 2

is that cos^2 x? or cos(2x)?

Answer 3

latter

Answer 4

I'd say rewrite cos(2x) as sin^2 x-cos^2 x

Answer 5

*cos^2 x-sin^2 x got the order wrong

Answer 6

no..
@SithsAndGiggles

Answer 7

wait cos(2x)=2cos^2 x-1 that should help you

Answer 8

\[\LARGE \int\limits\limits_{0}^{\frac{32 \pi}{3}}\sqrt{\cos^2x}~dx=\LARGE \int\limits\limits_{0}^{\frac{32 \pi}{3}}\|cosx|~dx\]

Answer 9

worry first about getting the integral down before you get the abs function etc right, since the cos function is periodic

Answer 10

The period is PI right?

Answer 11

you said it's sqrt(1-cos(2x)) correct?

Answer 12

Break it up into the "negative" and positive parts.

Answer 13

I applied cos2x=2 cos^2x - 1