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

Question

6153 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-04T00:28:18

$\frac{1}{2} e^{sin 2 x} + C$ $1/2e^(sin2x)+C$

This is the same as asking:

$\int (cos 2 x) e^{sin 2 x} d x$ $int(cos2x)e^(sin2x)dx$

Let $u = sin 2 x$ $u=sin2x$ . Differentiating this (and remembering the chain rule) gives us $d u = 2 cos 2 x . d x$ $du=2cos2xcolor(white).dx$ .

We have a factor of $2 cos 2 x$ $2cos2x$ , so modify the integral to make $2 cos 2 x$ $2cos2x$ present:

$= \frac{1}{2} \int (2 cos 2 x) e^{sin 2 x} d x$ $=1/2int(2cos2x)e^(sin2x)dx$

We now have our $u$ $u$ and $d u$ $du$ terms primed and ready for substitution:

$= \frac{1}{2} \int e^{u} d u$ $=1/2inte^udu$

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

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

$= \frac{1}{2} e^{sin 2 x} + C$ $=1/2e^(sin2x)+C$

We can check this by taking the derivative.

How do you find the antiderivative of (cos 2x)e^(sin2x)(cos2x)esin2x(cos 2x)e^(sin2x)?