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.

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