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

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Answer 1 · 2016-12-19T03:56:24

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

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#

Answer 2 · 2016-12-19T09:52:32

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

you can integrate this directly.

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

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