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

Jun 26, 2016

Use a $u$-substitution to find $\int {e}^{\sin} x \cdot \cos x \mathrm{dx} = {e}^{\sin} x + C$.

#### Explanation:

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

Let $u = \sin x \to \frac{\mathrm{du}}{\mathrm{dx}} = \cos x \to \mathrm{du} = \cos x \mathrm{dx}$

$\int {e}^{\sin} x \cdot \cos x \mathrm{dx}$ becomes:
$\int {e}^{u} \mathrm{du}$

This integral evaluates to ${e}^{u} + C$ (because the derivative of ${e}^{u}$ is ${e}^{u}$). But $u = \sin x$, so:
$\int {e}^{\sin} x \cdot \cos x \mathrm{dx} = \int {e}^{u} \mathrm{du} = {e}^{u} + C = {e}^{\sin} x + C$