Question

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

Answer 1

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

Explanation:

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`