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

Dec 4, 2016

#### Answer:

$\frac{1}{2} {e}^{\sin 2 x} + C$

#### Explanation:

This is the same as asking:

$\int \left(\cos 2 x\right) {e}^{\sin 2 x} \mathrm{dx}$

Let $u = \sin 2 x$. Differentiating this (and remembering the chain rule) gives us $\mathrm{du} = 2 \cos 2 x \textcolor{w h i t e}{.} \mathrm{dx}$.

We have a factor of $2 \cos 2 x$, so modify the integral to make $2 \cos 2 x$ present:

$= \frac{1}{2} \int \left(2 \cos 2 x\right) {e}^{\sin 2 x} \mathrm{dx}$

We now have our $u$ and $\mathrm{du}$ terms primed and ready for substitution:

$= \frac{1}{2} \int {e}^{u} \mathrm{du}$

The integral of ${e}^{u}$ is itself, since its derivative is itself:

$= \frac{1}{2} {e}^{u} + C$

$= \frac{1}{2} {e}^{\sin 2 x} + C$

We can check this by taking the derivative.