# Thread: Trig integrals

1. ## Trig integrals

Find the indefinite integral:

h(u) = sin^2(1/12u)

This is what I have:

sin^2(1/12u) = 1/2( 1 - cos( 1/6u))

sin^2(1/12u) = 1/2 -1/2 cos(1/6u) dx

= 1/12u-1/4sin(1/6) +c

I am lost, please help!

2. Originally Posted by mark090480
Find the indefinite integral:

h(u) = sin^2(1/12u)
Is "1/12u" supposed to be "1/(12u)" or "(1/12)u"?

I will assume you meant "(1/12)u".

This is what I have:

sin^2(1/12u) = 1/2( 1 - cos( 1/6u))

sin^2(1/12u) = 1/2 -1/2 cos(1/6u) dx

= 1/12u-1/4sin(1/6) +c
Please don't just write "=" like that. That means that what you are writing now is equal to what you had on the previous line. And that is not true.

I am lost, please help!
Why "(1/12)u"? The anti-derivative of (1/2) is (1/2)u. How did you get the "1/4" in front of "cos((1/6)u"?

3. No, I mean (1/12u) or (0.8333333u)

Thanks,

-Mark.