How do you integrate e^(sinx) cosx dx?

Dec 19, 2016

${e}^{\sin \left(x\right)} + C$

Explanation:

You can solve the integral using a u-substitution

Let $u = \sin \left(x\right)$

Differentiating we get

$\mathrm{du} = \cos \left(x\right) \mathrm{dx}$

Make the subtitution

$\int {e}^{u} \mathrm{du}$

integrating we get ${e}^{u}$

Now back substitute for $u$

${e}^{\sin \left(x\right)} + C$

Dec 19, 2016

if you recognise the result
$\text{ "d/(dx)(e^f(x))=f'(x)e^((f(x))" }$

you can integrate this directly.

Explanation:

$\text{ "d/(dx)(e^sinx)=cosxe^(sinx)" }$

so $\int \cos x {e}^{\sin x} \mathrm{dx} \text{ } = {e}^{\sin} x + C$