Integrating e^(2t)(25sint+20cost)dt using Integration by Parts

  • Thread starter mshiddensecret
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In summary, the conversation is about integrating a given expression using integration by parts. The person is unsure of what to set for u and v and ended up using Wolfram Alpha for the solution. The solution is e^(2t)(14sint+3cost)=C. The expert suggests splitting the integral into two parts and using different substitutions for u each time integration by parts is applied.
  • #1
mshiddensecret
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Homework Statement



Integrate: (e^(2t))(25sint+20cost)dt

Homework Equations



Integration by parts twice

The Attempt at a Solution



I don't know what to set for u and v.

I ended up wolfram alpha it but I want to know how to get there.

The solution is:

e^(2t)(14sint+3cost)=C
 
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  • #2
mshiddensecret said:

Homework Statement



Integrate: (e^(2t))(25sint+20cost)dt

Homework Equations



Integration by parts twice

The Attempt at a Solution



I don't know what to set for u and v.

I ended up wolfram alpha it but I want to know how to get there.

The solution is:

e^(2t)(14sint+3cost)=C
You could split the integral into two parts and work each one separately. I would start off with u1 = 25sin(t) for the first integral and u2 = 20cos(t) for the second one. Do similar substitutions the second time you apply integration by parts.
 
  • #3
mshiddensecret said:
e^(2t)(14sint+3cost)=C
Should be + C, right?
 

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