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anonymous
 2 years ago
How to find the indefinite integral of (e^t)(1+3sin(t))dt?
anonymous
 2 years ago
How to find the indefinite integral of (e^t)(1+3sin(t))dt?

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, you could. There's an answer for this.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0integral of e^t (1+3sin(t)) dt Expand it so e^t (1+3sin(t)) dt = (e^t + 3e^t sin(t)) dt then integrate each term and remove/factor constants so integral of e^t dt + 3 integral of e^t sin(t) dt therefore the integral of e^t dt = e^t, however for the integral: 3 integral of e^t sin(t) dt use the formula : integral of exponent ( alpha (t)) sin ( beta (t)) dt = exponent (alpha (t)) (beta cos ((beta)(t)) + alpha sin ((beta) (t)))/ alpha^2 + beta^2 after using the formula and integrating: 3 integral of e^t sin(t) dt we get : 3/2 e^t sin(t)  3/2e^t cos(t) then combine that with the integral of e^t dt = ? Hope you understand that. If you don't, there's absolutely nothing I can do about it.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Well lets separate it to \( \int e^{t}dt +3\int e^{t}sin(t)dt\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Credit: A beautiful cheating site that works for everyone http://answers.yahoo.com/question/index?qid=20091127013306AAGvR7G

GoldPhenoix
 2 years ago
Best ResponseYou've already chosen the best response.0His question and the person who asked in answeryahoo are different. :

GoldPhenoix
 2 years ago
Best ResponseYou've already chosen the best response.0What is your link suppose to be, Realist?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Ok Then we get \(\large\frac{e^{t}}{t}+3(\large \frac{e^{t}}{2}(sintcost)\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0It's supposed to be opened. @GoldPhenoix

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Ok so for the first integral I used the basic rule which is \(\large \int e^{at}dt = \frac{e^{at}}{t}+C\) And for the second integral I used \( \large \int e^{at}sinbt dt = \frac{e^{at}}{a^2+b^2}(a*sinbt b*sinbt)+C\) Whooopsss I forgot the +C in my answer above

Loser66
 2 years ago
Best ResponseYou've already chosen the best response.0@swissgirl Sorry for my dummy, I don't get the second integral.Please, explain me

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0You are not a dummy :) Umm I just used the table of integrals otherwise it gets messy

Loser66
 2 years ago
Best ResponseYou've already chosen the best response.0can I know where does that table come from?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Ummm its on the back page of every calc book

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0If you would like I can photocopy the page and upload it but i bet u can find it online

Loser66
 2 years ago
Best ResponseYou've already chosen the best response.0ok, let me check. Since I take derivative of the answer, I don't get the integrand, but a mess. hihihi... May be because I made mistake at somewhere. let me redo. Thanks for response

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Not necessarily is there a mistake maybe its the simplification thats the issue. It happens to me all the time

Loser66
 2 years ago
Best ResponseYou've already chosen the best response.0hey, I got it from my book too. hihi. sorry.
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