How do you integrate #e^(cos3x)sin3xdx#?

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Answer 1 · 2015-04-03T14:04:47

This is almost, but not quite #int e^u du#. Try to make it of that form.

Let #u=cos 3x#. This makes #du = -3sin3x dx#.

Use whatever details of substitution you've been taught to get:

#int e^(cos3x) sin3x dx= -1/3 int e^u du = -1/3e^u +C#

#=-1/3 e^(cos3x) +C#.

Alternative Notation:
Let #g(x)=cos 3x#. This makes #g'(x) = -3sin3x#.

#int e^(cos3x) sin3x dx = -1/3 int e^g(x) g'(x)dx = -1/3 e^ g(x) +C#

#=-1/3 e^( cos3x) +C#