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 $g'(x) = -3sin3x$ .

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

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

How do you integrate e^(cos3x)sin3xdxe^(cos3x)sin3xdx?