How do you find the antiderivative of #(cos 2x)e^(sin2x)#?

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Answer 1 · 2016-12-04T00:28:18

#1/2e^(sin2x)+C#

This is the same as asking:

#int(cos2x)e^(sin2x)dx#

Let #u=sin2x#. Differentiating this (and remembering the chain rule) gives us #du=2cos2xcolor(white).dx#.

We have a factor of #2cos2x#, so modify the integral to make #2cos2x# present:

#=1/2int(2cos2x)e^(sin2x)dx#

We now have our #u# and #du# terms primed and ready for substitution:

#=1/2inte^udu#

The integral of #e^u# is itself, since its derivative is itself:

#=1/2e^u+C#

#=1/2e^(sin2x)+C#

We can check this by taking the derivative.