Question

∫ (e^(cos3x)) times (sin3xdx)

Answer 1

The derivative of cos3x is -sin3x. So we try the substitution:
\[ u=cos3x \\
du=-sin3x \ dx \]
This yields:
\[ \int{e^{cos3x}sin3x \ dx} = \int{e^u \ -du} = -e^u+C \]
Substituting back for x we have the solution
\[ -e^{cos3x}+C \]

Answer 2

That is e^u times (-du) in the middle step.

Answer 3

Ah, I'm tired. du is -3sin3x dx and not -sin3x.
The solution is therefore
\[ -\frac{1}{3}e^{cos3x}+C \]

Answer 4

Sorry for that, just ask if my answer became confusing.

Answer 5

Nope, explanation was great! thanks a ton