Original post by zakourad
Hey Guys im stuggling to find a way to integrate this question i keep ending up in an endless loop if attemptiong to integrate the bellow by parts.
e^xsinx
e^xsinx
You have a choice as to which part you make the "u" and which part the "v" in your IBP. One way takes you round in circles, the other works out after two applications.
You end up with I = something + "a constant" times I, where "I" is your original integral, and rearrange and solve for I.
Original post by zakourad
Hey Guys im stuggling to find a way to integrate this question i keep ending up in an endless loop if attemptiong to integrate the bellow by parts.
e^xsinx
e^xsinx
I think that if you make xsinx as the u, you won't be going round in circles..
Original post by zakourad
Hey Guys im stuggling to find a way to integrate this question i keep ending up in an endless loop if attemptiong to integrate the bellow by parts.
e^xsinx
e^xsinx
I'm assuming this is (e^x)(sinx) rather than e^(xsinx), which would be much more difficult.
Original post by ghostwalker
You have a choice as to which part you make the "u" and which part the "v" in your IBP. One way takes you round in circles, the other works out after two applications.
You end up with I = something + "a constant" times I, where "I" is your original integral, and rearrange and solve for I.
You end up with I = something + "a constant" times I, where "I" is your original integral, and rearrange and solve for I.
I think you can choose either function as your "u" to start with, but if you choose e^x as u in the 1st pass, you have to use it as u in the 2nd pass too, and similarly if you choose the trig function as your "u" 1st time.
Original post by lionelmessi19
I think that if you make xsinx as the u, you won't be going round in circles..
xsinx isn't a valid choice for the "u" here - IBP deals with the product of two functions and the two functions available are e^x and sin x (technically you could choose 1 as "u" and use (e^x)sinx as the dv/dx (or vice versa) but that won't really help here
Original post by davros
I think you can choose either function as your "u" to start with, but if you choose e^x as u in the 1st pass, you have to use it as u in the 2nd pass too, and similarly if you choose the trig function as your "u" 1st time.
My bad - I assumed without checking. PRSOM.
Original post by davros
I think you can choose either function as your "u" to start with, but if you choose e^x as u in the 1st pass, you have to use it as u in the 2nd pass too, and similarly if you choose the trig function as your "u" 1st time.
xsinx isn't a valid choice for the "u" here - IBP deals with the product of two functions and the two functions available are e^x and sin x (technically you could choose 1 as "u" and use (e^x)sinx as the dv/dx (or vice versa) but that won't really help here
xsinx isn't a valid choice for the "u" here - IBP deals with the product of two functions and the two functions available are e^x and sin x (technically you could choose 1 as "u" and use (e^x)sinx as the dv/dx (or vice versa) but that won't really help here
Yeah, you're right.. it doesn't work!
So how would you answer it then?Original post by lionelmessi19
Yeah, you're right.. it doesn't work! So how would you answer it then?
So how would you answer it then?As ghostwalker first suggested, you need two iterations of integration by parts, but you must choose the exponential as the first function both times or the trig function as the first function both times - if you "swap over" then you'll just go round in circles because you'll be differentiating something you've previously integrated, or vice versa.
If you let I = integral of (e^x)sinx and J = integral of (e^x)cosx then the first IBP will give you something like
I = f(x) + kJ
You can then integrate J by parts to get something like
J = g(x) + mI
where f(x), g(x) are some functions of x involving exponentials and trig, and k and m are some constants.
If you plug the 2nd equation into the first one, you then get something like
I = h(x) + pI (where h(x) is some combination of f(x), g(x) and constants; and p is a constant involving k and m)
which you can rearrange to solve for I
Original post by davros
As ghostwalker first suggested, you need two iterations of integration by parts, but you must choose the exponential as the first function both times or the trig function as the first function both times - if you "swap over" then you'll just go round in circles because you'll be differentiating something you've previously integrated, or vice versa.
If you let I = integral of (e^x)sinx and J = integral of (e^x)cosx then the first IBP will give you something like
I = f(x) + kJ
You can then integrate J by parts to get something like
J = g(x) + mI
where f(x), g(x) are some functions of x involving exponentials and trig, and k and m are some constants.
If you plug the 2nd equation into the first one, you then get something like
I = h(x) + pI (where h(x) is some combination of f(x), g(x) and constants; and p is a constant involving k and m)
which you can rearrange to solve for I
If you let I = integral of (e^x)sinx and J = integral of (e^x)cosx then the first IBP will give you something like
I = f(x) + kJ
You can then integrate J by parts to get something like
J = g(x) + mI
where f(x), g(x) are some functions of x involving exponentials and trig, and k and m are some constants.
If you plug the 2nd equation into the first one, you then get something like
I = h(x) + pI (where h(x) is some combination of f(x), g(x) and constants; and p is a constant involving k and m)
which you can rearrange to solve for I
Hmmm.. thanks for that! Nice way of explaining without disclosing the answer! Cheers
+1 rep
Original post by lionelmessi19
Hmmm.. thanks for that! Nice way of explaining without disclosing the answer! Cheers
+1 rep
+1 rep
As you can tell, I was trying very hard not to give away the exact answer
In this case, things won't be too complicated - all the functions that pop out of the various stages of IBP will look like e^xsinx or e^xcosx, and you can probably see that the constants that come out will also either be +1 or -1, so probably the "worst" thing that can happen at the end is you end up with a factor of 2 or 1/2 to divide by at some stage!
It gets a bit messier when you have random things like to integrate because then you get lots of constants being multiplied (when you differentiate) or divided by (when you integrate) that you have to keep track of, and it's easy to slip up part way through the process!
Just integrate until you get the to the point where you are subtracting the integral you are trying to find, then plus it over so you get 2 x the integral you are trying to find and then divide by 2.
For example if you are trying to find the integral 'I' and you get to the stage where
I = g(x) - I
Then
2I = g(x)
Therefore
I = g(x)/2
Posted from TSR Mobile
For example if you are trying to find the integral 'I' and you get to the stage where
I = g(x) - I
Then
2I = g(x)
Therefore
I = g(x)/2
Posted from TSR Mobile
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