Kirill Z. answered • 09/16/13

Physics, math tutor with great knowledge and teaching skills

You can do it by applying the integration by parts.

∫e

^{3x}cos(2x)dx=(½)e^{3x}sin(2x)-(3/2)∫e^{3x}sin(2x)dx=(½)e^{3x}sin(2x)-(¾)e^{3x}(-cos(2x))-(9/4)∫e^{3x}cos(2x)dxNow let I=∫e

^{3x}cos(2x)dx; Then we have an equation:I=(½)e

^{3x}sin(2x)+(¾)e^{3x}cos(2x)-(9/4)I;Solving for I yields:

I=1/13*[2e

^{3x}sin(2x)+3e^{3x}cos(2x)]=e^{3x}/13*[2sin(2x)+3cos(2x)]
James R.

09/16/13