I cant seem to figure out what the integral of this is correctly. If anyone can help it would be appreciated as i need the integral for part of a project.

Note: The entire function is squared, not just the (x-4).

Thanks

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- May 28th 2011, 01:26 PMbonker98Help with integral of (4sin(x-4))^2
I cant seem to figure out what the integral of this is correctly. If anyone can help it would be appreciated as i need the integral for part of a project.

Note: The entire function is squared, not just the (x-4).

Thanks - May 28th 2011, 01:28 PMTed
Well,

It could be written as : $\displaystyle 16 \int sin^2(x-4) \, dx$. Right?

Substitute $\displaystyle u=x-4~.$ then use a trigonometry identity. Tell us what will you get. - May 28th 2011, 01:31 PMe^(i*pi)
Since $\displaystyle (4\sin(x-4))^2 = 16\sin^2(x-4)$ we can pull the 16 out front: $\displaystyle 16 \int \sin^2(x-4)$ but to make it easier I'm going to write is at $\displaystyle 8 \int 2\sin^2(x-4)$

You can then use a double angle identity: *edited out*