# help with integral

• Feb 18th 2007, 08:29 PM
rcmango
help with integral
heres the problem: http://img186.imageshack.us/img186/4328/untitledil5.jpg

i marked where i'm having trouble.

I don't see how this was integrated.
can someone break it down further, for the part i marked.

thanks
• Feb 18th 2007, 08:43 PM
ThePerfectHacker
Quote:

Originally Posted by rcmango
heres the problem: http://img186.imageshack.us/img186/4328/untitledil5.jpg

i marked where i'm having trouble.

I don't see how this was integrated.
can someone break it down further, for the part i marked.

thanks

The integral of 9/2 is 9/2 x.

The integral of 6*sin x=-6cos x.

The integral of 2*sin^2 x is tricker.
You can write it using the half-angle identity to get,
2*sin^2 x = 1-cos 2x and then that easily integrates.

And the result is the sum of those.
• Feb 18th 2007, 10:19 PM
rcmango
[QUOTE=1-cos 2x and then that easily integrates.QUOTE]

so then using the trig identity, sinx^2 + cosx^2 = 1

1 - cos2x is the integral.

i'm going to guess what the derivative is, may need help, tried using mathematica.

hmm not sure why x is dived by 2, wheres the 1/4 coming from? .. x/2 - 1/4sin(2x) + c

thanks for all the help so far.
• Feb 19th 2007, 08:34 AM
ThePerfectHacker
Quote:

so then using the trig identity, sinx^2 + cosx^2 = 1

1 - cos2x is the integral.

i'm going to guess what the derivative is, may need help, tried using mathematica.

hmm not sure why x is dived by 2, wheres the 1/4 coming from? .. x/2 - 1/4sin(2x) + c

thanks for all the help so far.
The half-angle identity is:

sin^2(x/2)=(1/2)(1-cos x)

cos^2(x/2)=(1/2)(1+cos x)