Hi,
Wondering if someone could point me in the right direction for this question. It is an assignment question, so I don't want the answer, just some help and point in the right direction. Heres what I have so far:
Find indefinite integral of: h(u) = sin^2 ( 1/6 u )
Now I'm presuming I need to use the double angle formula
cos(2x) = 1 - 2sin^2 x
to which I have:
sin^2 x = 1/2 (1 - cos (2x))
sin^2 (1/6 u ) = 1/2 (1 - cos (2 (1/6 u))
Therefore I get the answer
integral sin^2(1/6u) = integral (1/2 - 1/2 cos (2 (1/6 u )))
=1/2x - 3/2 sin (1/3 u ) + c
Is that anywhere close?