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anonymous
 5 years ago
∫[e^(3t)sin(3t),dt]= ;limit(0,π)
anonymous
 5 years ago
∫[e^(3t)sin(3t),dt]= ;limit(0,π)

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0you will definitely need to use integration by parts

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0if you know the product rule, you can derive the formula for integration by parts

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0if you don't know the product rule, you can use the definition of derivative to find the formual for the product rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Integrate by parts, setting e^(3x) equal to u and the trig function equal to v' each time. The 3's will cancel out every time, and you'll end up in the third term with the original integral; add it to the left and divide by two, you'll end up getting\[e^{3t}*1/6*(\sin(3t)\cos(3t)).\] Evaluate it at the limits and out pops the value. :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is the exact ans I got when I integrated it, QuantumModulus.....But my final ans was 1/6(1+e^3pi)
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