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?

That's on the right track. I've not gone through it to check exact numbers etc, but the logic is reasonably correct.

Cheers

I checked it, nothing wrong at all.

"integral sin^2(1/6u) = integral (1/2 - 1/2 cos (2 (1/6 u )))

 

=1/2x - 3/2 sin (1/3 u ) + c

"

 

you added in a variable x, just nitpicking though, I'm sure you meant to say u

yep fine. BE Careful with changing your letters.

(in math examiner mode ) they need to see what you're integrating with respect to.

ie in this case is it dx or du?

only coving your back!

hehe, cheers. Its already been handed in, using x instead of u. Knowing my tutor I'll loose that mark for changing the letters, ah well.

Thanks for your help