I posted this below but had no response. I would appreciate help with details please.How do I find the indefinite integral of h(u)=sin^2(1/5u) this is one fifth u. Does it involve double angle formulas? Thanks.

Answer 1

start with the identity
cos 2x = 1 - 2sin^2 x
2sin^2 x = 1 - cos 2x
sin^2 x = 1/2 - (1/2)cos 2x

then sin^2 (u/5) = 1/2 - (1/2)cos (2u/5)

and the integral of that is
(1/2)u - (1/2)sin(2u/5) (5/2)
= (1/2)u - (5/4)sin(2u/5) + c

Answer 2

sin^2(u/5) = (1/2)[1 - cos(2u/5)]

The indefinite integral of that with respect to u is

u/2 - (1/2)*sin(2u/5)*(5/2)
= u/2 - (5/4)*sin(2u/5)