Se^(3x)cos(2x)dx S=Integral

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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?

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)dx

Now 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)]

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