How do you find the antiderivative of $e^(sinx)*cosx$ ?

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Answer 1 · 2016-06-26T02:35:30

Use a $u$ -substitution to find $inte^sinx*cosxdx=e^sinx+C$ .

Notice that the derivative of $sinx$ is $cosx$ , and since these appear in the same integral, this problem is solved with a $u$ -substitution.

Let $u=sinx->(du)/(dx)=cosx->du=cosxdx$

$inte^sinx*cosxdx$ becomes:
$inte^udu$

This integral evaluates to $e^u+C$ (because the derivative of $e^u$ is $e^u$ ). But $u=sinx$ , so:
$inte^sinx*cosxdx=inte^udu=e^u+C=e^sinx+C$

How do you find the antiderivative of e^(sinx)*cosxe^(sinx)*cosx?