Integration help

[Prometheus83]

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?

[Dave]

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

[Prometheus83]

Cheers

[Pat Says]

I checked it, nothing wrong at all.

[CPL.Luke]

"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

[kingjewel1]

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!

[Prometheus83]

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