How do you evaluate the integral #int e^xcosx#?

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Answer 1 · 2017-01-18T15:48:07

#int e^xcosxdx =e^x((cosx + sinx)/2)+C#

We can integrate by parts using the differential:

#d(e^x) = e^xdx#

so that:

#int e^xcosxdx = int cosx d(e^x) = e^xcosx - int e^xd(cosx)= e^xcosx + int e^xsinxdx#

We can solve this last integrate by parts again:

#int e^xsinxdx = int sinx d(e^x) = e^xsinx - int e^xd(sinx) = e^xsinx - int e^xcosxdx#

so that we have:

#int e^xcosxdx =e^xcosx + e^xsinx - int e^xcosxdx#

Now the same integral is appearing at both members and we can solve for it:

#2int e^xcosxdx =e^xcosx + e^xsinx#

#int e^xcosxdx =e^x((cosx + sinx)/2)+C#