I just got back from my diff eq exam.
One of the problems was y'+y*(tan(x)) = 8*sin(x^3)
At a first glance, it looks like a simple first order linear diff eq. But after getting an integrating factor of sec(x), how do I find the integral of 8*sin(x^3)cos(x)?
Wolfram alpha seems to be returning a cryptic answer.
Pardon my ineptness at math. I had it confused with the sin^2(x) and [sin(x)]^2 identity. But yes, it's [sin(x)]^3
How do you do the U substitution? With sin^3(x)dx/cos(x) let cos(x)=u, -du= sinxdx, this doesn't seem to work since there will still be sin^2x left
Let u = sin(x), du=cosxdx... I can't get that to work since it's dx/cos(x).
MUCHO thanks for the help