Calculus - Integration by Parts
posted by Emma .
Please help me solve this integration problem. I don't understand how to do it. Thank you!
I do not believe you can do this in one integration-by-parts step. You must use the method twice.
Let y = 2x and dx = dy/2, to simplify the problem to
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