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How to find the indefinite integral of (e^-t)(1+3sin(t))dt?

Mathematics
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Other answers:

integral 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.
Well lets separate it to \( \int e^{-t}dt +3\int e^{-t}sin(t)dt\)
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Ok Then we get \(\large\frac{-e^{-t}}{t}+3(\large \frac{e^{-t}}{2}(-sint-cost)\)
It's supposed to be opened. @GoldPhenoix
Ok 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
Thanks a lot.
@swissgirl Sorry for my dummy, I don't get the second integral.Please, explain me
You are not a dummy :) Umm I just used the table of integrals otherwise it gets messy
can I know where does that table come from?
Ummm its on the back page of every calc book
If you would like I can photocopy the page and upload it but i bet u can find it online
ok, 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
Not necessarily is there a mistake maybe its the simplification thats the issue. It happens to me all the time
hey, I got it from my book too. hihi. sorry.

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