The question h(u)=sin^2(8u)
I used the double angle formulas and the answer I got is
1/32sin(16u)+1/2u+c
Is this right
Ok i can't see where I've gone wrong in my workings out so I will post them and could you have a look please (can't find a theta sign so i'll type
it)
sin^2(8u)
where
sin^2=1-cos^2theta here theta=8u
so
sin^2(8u)=1-cos^2(8u)
sin^2(8u)=1-cos^2(16u)
sin^2(8u)=1/2(cos(16u)+1)
sin^2(8u)=(1/2cos(16u)+1/2)=1/32sin(16u)-1/2u+c
Thank you