Integration problem. College Calc II?

Question

integral of e^(2x) * cosx dx. says to integrate by parts where you have a u and dv and substitute in. thanks!

Answer 1

Lol, i remember doing this problem too, but i kinda forgot some of my BC stuff.

u= e^2x , du= 2e^2x

v = sinx, dv = cosx

INT{udv} = uv - INT{vdu}

So u get this:

- INT{e^2x*cosxdx} = e^2x * sinx - INT{sinx * 2e^2x}

- So your gonna have to integrate INT{sinx * 2e^2x} by parts as well.

    - u= 2e^2x, du = 4e^2x

      v= - cosx, dv = sinx

     INT{sinx * 2e^2x} = 2e^2x * (-cosx) - INT{-cosx * 4e^2x}

So u get this  

  INT{e^2x*cosxdx} = e^2x * sinx - 2e^2x * (-cosx) - INT{-cosx * 4e^2x}

which =  

INT{e^2x*cosxdx}=e^2x * sinx - 2e^2x * (-cosx) + 4 * INT{cosx * e^2x}

Subtract the 4 * INT{cosx * e^2x} from the right side to the left and you get this:

    - 3* INT{e^2x*cosxdx} = e^2x * sinx - 2e^2x * (-cosx)



- So then you divide by (-3) to both sides and you get:

      INT{e^2x*cosxdx} = [e^2x * sinx - 2e^2x * (-cosx) ] / -3

Umm hopefully this is right, if not, well i tried, im somewhat sketchy in my math rite now. Good luck.  

Answer 2

Integrate by parts twice.

∫e^(2x)cos(x) dx

u=e^(2x) ; du = 2e^(2x) dx

dv = cos(x)dx ; v = sin(x) + c

∫udv = uv - ∫vdu

∫e^(2x)cos(x) dx = sin(x)e^(2x) - 2∫e^(2x)sin(x)dx + c

∫e^(2x)sin(x)dx

u=e^(2x) ; du = 2e^(2x) dx

dv = sin(x)dx ; v = -cos(x) + c

∫e^(2x)sin(x)dx = -e^(2x)cos(x) + 2∫e^(2x)cos(x)dx + c

∫e^(2x)cos(x) dx = sin(x)e^(2x) - 2[-e^(2x)cos(x) + 2∫e^(2x)cos(x)dx] + c

∫e^(2x)cos(x) dx = sin(x)e^(2x) + 2e^(2x)cos(x) - 4∫e^(2x)cos(x)dx + c

5∫e^(2x)cos(x) dx = sin(x)e^(2x) + 2e^(2x)cos(x) + c

∫e^(2x)cos(x) dx = (1/5)e^(2x)[sin(x) + 2cos(x)] + c

Answer: (1/5)e^(2x)[sin(x) + 2cos(x)] + c