Integrate cosx.e^x?

Answer 1

you have to integrate by parts TWICE. The first time you do it, you get a sin(x)e^x term, you integrate again to get a cos(x)e^x back (make sure you are defining the right parts as "u" and "dv"). It's hard to explain with words, but after that, you will see that you can solve for the integral of cos(x)e^x

Answer 2

Two ways to do this:

1) Integrate by parts twice. For first one, u=e^x, dv=cos(x)dx. For
the second, u=e^x and dv=sin(x) dx. When you are finished, you can solve back for the original integral.

2) write e^x cos(x) as the real part of e^[(1+i)x].
This can be integrated as an exponential and then converted back into trig functions using e^(ix)=cos(x)+i sin(x). You will have to rationalize the denominator first, though.

Answer 3

Integral e^(au)*cos(bu) du = [ (e^(au))/(a^2 +b^2)]*[a cos(bu) + b sin(bu)] + C

looks complicated but if you have x=u & a,b=1 you get:

Integral of (e^x)*cos(x) dx = [ (e^(x))/(1^2 +1^2)]*[1 cos(x) + 1 sin(x)] + C

which = [ ( e^x) /2]* [cos(x) + sin (x)]

=[(e^x*cos(x))/2] + [(e^x*sin(x))/2]