Calculus  Integration by Parts
posted by Emma .
Please help me solve this integration problem. I don't understand how to do it. Thank you!
∫e^(2x)*sin(2x)dx

I do not believe you can do this in one integrationbyparts step. You must use the method twice.
Let y = 2x and dx = dy/2, to simplify the problem to
(1/2)∫e^y*sin(y)dy
Next, let e^y = u and dv = siny dy
du = e^y dy and v = cos y
(1/2)∫e^y*sin(y)dy = (1/2)*uv (1/2)∫v*du
= (1/2)[u cos y +∫e^y cosy dy]
Now you must apply integration by parts a second time on the
+∫e^y cosy dy term
This will give you an equation for
∫e^y*sin(y)dy that involves explicit functions of y.
When you are all done, substitute 2x for y
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