How do you find the antiderivative of #(cos 2x)e^(sin2x)#?
1 Answer
Dec 4, 2016
Explanation:
This is the same as asking:
#int(cos2x)e^(sin2x)dx#
Let
We have a factor of
#=1/2int(2cos2x)e^(sin2x)dx#
We now have our
#=1/2inte^udu#
The integral of
#=1/2e^u+C#
#=1/2e^(sin2x)+C#
We can check this by taking the derivative.