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

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The integral of e^(2t)(25sin(t) + 20cos(t))dt can be solved using integration by parts, applied twice. A suggested approach is to split the integral into two parts, using u1 = 25sin(t) for the first and u2 = 20cos(t) for the second. The final solution is e^(2t)(14sin(t) + 3cos(t)) + C. Clarification was sought regarding the inclusion of the constant C in the solution. The discussion emphasizes the importance of proper variable substitution in integration by parts.
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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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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.
 
mshiddensecret said:
e^(2t)(14sint+3cost)=C
Should be + C, right?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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