Evaluate the integral... Se^(3x)cos(2x)dx S=Integral

Question

Se^(3x)cos(2x)dx  S=Integral

Answer 1

You can do it by applying the integration by parts.

 

∫e^3xcos(2x)dx=(½)e^3xsin(2x)-(3/2)∫e^3xsin(2x)dx=(½)e^3xsin(2x)-(¾)e^3x(-cos(2x))-(9/4)∫e^3xcos(2x)dx

 

Now let I=∫e^3xcos(2x)dx; Then we have an equation:

 

I=(½)e^3xsin(2x)+(¾)e^3xcos(2x)-(9/4)I;

 

Solving for I yields:

 

I=1/13*[2e^3xsin(2x)+3e^3xcos(2x)]=e^3x/13*[2sin(2x)+3cos(2x)]

 

Answer 2

The first step is to convert the cosine function to the exponential form using Euler's formula.

 

Cos(2x) = 1/2 * [e^(i2x)-e^(-i2x)]

 

The next step is to distribute the outside exponential that is 1/2*[e^(3x+i2x)-e^(3x-i2x)].

 

Now the integral becomes a very simple u and v substitution integral. That is let u = (3+i2)x and v = (3-i2)x therefore du = (3+i2)dx and dv = (3-i2)dx.

 

Is this far enough help or do you need the next steps?