Question

evaluate integral sign e^(x)sinx dx

Answer 1

oh the classic!

Answer 2

by parts? or is there a nicer way

Answer 3

\[\int\limits_{}^{} e^x \sin x dx\]

Answer 4

by parts, and you will wind up having to add the last integral to the first to break the cycle

Answer 5

it's a classic problem of... I forget what it's called
recursive integrals or something...

Answer 6

the answer for this problem is \[e^x (sinx-cosx)/2 + C\]

can u please show me whats the complete solution on this ?

Answer 7

\[u=e^x\]\[dv=\sin x\]\[\int e^x\sin xdx=-e^x\cos x+\int e^x\cos x\]\[=-e^x\cos x+e^x\sin x-\int e^x\sin xdx\]with me so fat?

Answer 8

so far*
lol

Answer 9

lol

Answer 10

integration by parts (thought i'd post it just in case) :

\[\int\limits{\frac{dv}{dx} \times u} \text{ } dx = u \times v - \int\limits{\frac{du}{dx} \times v} \text{ }dx\]

Answer 11

how do you identify a certain integral problem if its by part or something did anybody know ?just for a simple glance

Answer 12

in general when you have\[\int x^n\sin xdx\] or any other trig function, integration by parts is the way to go
in this case it's a little different, but this is a very well-known problem, so once you have seen how to do it you will recognize the way integrals like\[\int a^x\cos xdx\]go, and how they always involve integration by parts

Answer 13

...I shouldn't say always, I'm sure there are some exceptions out there

Answer 14

do you have any recommended PDF which surely give a sure fire to understand immediately integral probs? can I have it then?

Answer 15

I always refer people here:
http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx
this is a very good set of notes for many topics in first-year uni math

Answer 16

@TuringTest thank you so much....

Answer 17

you're very welcome :)
anyway we haven't finished our problem...
\[\int e^x\sin xdx=-e^x\cos x+e^x\sin x-\int e^x\sin xdx\]it looks like we are stuck here, since the last integral is back where we started, but now here is the trick:
let\[\int e^x\sin xdx=I\]we then have\[I=-e^x\cos x+e^x\sin x-I\]which can be solved for I....

Answer 18

\[2I=e^x(\sin x-\cos x)\implies I=\frac{e^x(\sin x-\cos x)}2+C\]and subbing back in for \(I\) we now see that\[\int e^x\sin xdx=\frac{e^x(\sin x-\cos x)}2+C\]and there you have it :)

Answer 19

thank you so much @TuringTest

Answer 20

again, you're very welcome :)