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

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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Let;

`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`

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