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

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2 Answers

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.
 
∫e3xcos(2x)dx=(½)e3xsin(2x)-(3/2)∫e3xsin(2x)dx=(½)e3xsin(2x)-(¾)e3x(-cos(2x))-(9/4)∫e3xcos(2x)dx
 
Now let I=∫e3xcos(2x)dx; Then we have an equation:
 
I=(½)e3xsin(2x)+(¾)e3xcos(2x)-(9/4)I;
 
Solving for I yields:
 
I=1/13*[2e3xsin(2x)+3e3xcos(2x)]=e3x/13*[2sin(2x)+3cos(2x)]
 

Comments

I do get the e^(3x)/13 in front but I have a much more simple expression in the brackets.
I checked with Mathematica. My answer is correct. ;-)
After relooking at my method, it does simplify to the answer you give. It just gives the student two different methods for solving the same problem.