How do you evaluate the integral int e^xcosxexcosx?

1 Answer
Jan 18, 2017

int e^xcosxdx =e^x((cosx + sinx)/2)+Cexcosxdx=ex(cosx+sinx2)+C

Explanation:

We can integrate by parts using the differential:

d(e^x) = e^xdxd(ex)=exdx

so that:

int e^xcosxdx = int cosx d(e^x) = e^xcosx - int e^xd(cosx)= e^xcosx + int e^xsinxdxexcosxdx=cosxd(ex)=excosxexd(cosx)=excosx+exsinxdx

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^xcosxdxexsinxdx=sinxd(ex)=exsinxexd(sinx)=exsinxexcosxdx

so that we have:

int e^xcosxdx =e^xcosx + e^xsinx - int e^xcosxdxexcosxdx=excosx+exsinxexcosxdx

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

2int e^xcosxdx =e^xcosx + e^xsinx2excosxdx=excosx+exsinx

int e^xcosxdx =e^x((cosx + sinx)/2)+Cexcosxdx=ex(cosx+sinx2)+C