How do you find the antiderivative of $e^{sin x} \cdot cos x$ $e^(sinx)*cosx$ ?

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How do you find the antiderivative of $e^{sin x} \cdot cos x$ $e^(sinx)*cosx$ ?

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Answer 1 · 2016-06-26T02:35:30

Use a $u$ $u$ -substitution to find $\int e^{sin x} \cdot cos x d x = e^{sin x} + C$ $inte^sinx*cosxdx=e^sinx+C$ .

Explanation:

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

Let $u = sin x \to \frac{d u}{d x} = cos x \to d u = cos x d x$ $u=sinx->(du)/(dx)=cosx->du=cosxdx$

$\int e^{sin x} \cdot cos x d x$ $inte^sinx*cosxdx$ becomes:
$\int e^{u} d u$ $inte^udu$

This integral evaluates to $e^{u} + C$ $e^u+C$ (because the derivative of $e^{u}$ $e^u$ is $e^{u}$ $e^u$ ). But $u = sin x$ $u=sinx$ , so:
$\int e^{sin x} \cdot cos x d x = \int e^{u} d u = e^{u} + C = e^{sin x} + C$ $inte^sinx*cosxdx=inte^udu=e^u+C=e^sinx+C$

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