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

2 Answers
Dec 19, 2016

Answer:

#e^(sin(x))+C#

Explanation:

You can solve the integral using a u-substitution

Let #u=sin(x)#

Differentiating we get

#du=cos(x)dx#

Make the subtitution

#int e^udu#

integrating we get #e^u#

Now back substitute for #u#

#e^(sin(x))+C#

Dec 19, 2016

Answer:

if you recognise the result
#" "d/(dx)(e^f(x))=f'(x)e^((f(x))" "#

you can integrate this directly.

Explanation:

#" "d/(dx)(e^sinx)=cosxe^(sinx)" "#

so #intcosxe^(sinx)dx" "=e^sinx+C#