Use integration by parts to determine:   `int` e^x cos 2x dx

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jeew-m eNotes educator| Certified Educator


`U = cos2x`

`V = e^x`


`dU = -2sin2xdx`

`dV = e^xdx`


using integration by parts;

`intUdV = UV-intVdU`

`intcos2xe^xdx = (cos2x*e^x)-inte^x(-2sin2x)dx`

`intcos2xe^xdx = (cos2x*e^x)+2inte^xsin2xdx`


Let us consider the part inte^xsin2xdx

When you apply integral by parts for this you will get;

`P = sin2x`

`Q = e^x`


`dP = 2cos2xdx`

`dQ = e^xdx`


`intsin2xdQ = PQ-inte^xdP`

`inte^xsin2x = (e^xsin2x)-2inte^xcos2xdx`


`intcos2xe^xdx = (cos2x*e^x)+2inte^xsin2xdx`

`intcos2xe^xdx = (cos2x*e^x)+2((e^xsin2x)-2inte^xcos2xdx)`

`intcos2xe^xdx = (cos2x*e^x)+2(e^xsin2x)-4inte^xcos2xdx`


`intcos2xe^xdx+4inte^xcos2xdx = e^x(cos2x+2sin2x)`

`5inte^xcos2xdx = e^x(cos2x+2sin2x)`

`inte^xcos2xdx = (e^x(cos2x+2sin2x))/5`


So the answer is;

`inte^xcos2xdx = (e^x(cos2x+2sin2x))/5`