Integrate e^((-3x+2sinx)/6)
Thanks. It could be. Any indications would be helpful for me to investigate.
Wait. That 1/6 is part of the exponent?
Yes
You could have just left it and put 1/e^6 outside the integral.
exp ((-3x +2sinx)\div6)
Oh yah.
e^1/6 is the integral?
No. That's just a constant you can pull out of the integral
That still looks like a tough one.
yep haha
\[-1/2\exp ^{2sinx-3x}\]
That where i get to but can't find out what to do with 2sinx
Wolfram Alpha - http://www.wolframalpha.com/input/?i=integrate+e%5E%282*sinx+-+3*x%29
daaamn. if wolfram cant do it then i'm not going to try
Are you supposed to get the indefinite integral or maybe an approximation of a definite integral?
I think that is simpler than it seems but I have missed a trick?
I need to get an indefinite integral
Ok. I may have found the issue. Does it make a difference if I have (2cosx-3)
with it
This function as written does not have a closed form answer. Where did you get this integral?
this looks the same: http://openstudy.com/study#/updates/4faf997ee4b059b524fa839e
so was i late?
I need to find the general solution to the following in implicit form \[dy/dx = (2cosx -3) \exp ^{(1/6)(-3x+2sinx))} y ^{5/6} (y>0)\]
aha! that is easier i think, we have a substitution.
Of course that is easier. Why did you hide ( 2 cos x -3?
\[\int\limits{y^{\frac{-5}{6}}}dy = \int\limits{(2cosx - 3)e^{\frac{1}{6} (-3x + 2sinx)}}dx\] u = -3x + 2sinx
And I use the substitution method?
yes.
@WonHungLo i think we were wrong when we said we could take that factor out earlier.
Thanks everybody. I can now investigate all this and see how we get to where we got to. I will need time but I am learning! YES!
@hmidante, please write down the whole question and not one part of it.
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