Posted by Tracy on Sunday, June 10, 2007 at 11:10pm.
so we are doing integrals and I have this question on my assignment and I can't seem to get it, because we have the trig substituion rules, but the number isn't even so its not a perfect square and I jsut cant get it, so any help would be greatly appreciated. The question is :
integral of du/ (u*sqrt(4-u^2))
Thanks again
Use the concept of U substitution!
--Becky Meyers, Harvard Class of 2011
We have to use Trig substition before we can use U substution because it won't work the other way, I tried, I know what the answer is supposed to be I just can't get it.
integral of du/ (u*sqrt(4-u^2))
Substitute u = 1/t. Then
du = -dt/t^2
u*sqrt[4-u^2] = 1/t sqrt[4-1/t^2]
du/ (u*sqrt(4-u^2)) =
-dt/t^2 * t/sqrt[4-1/t^2] =
-dt/[t sqrt(4-1/t^2)] =
-dt/sqrt[4 t^2 - 1]
So, you should do a hyperbolic substitution:
put t = 1/2 cosh(y)
then
dt = 1/2 sinh(y)
sqrt[4 t^2 - 1] = sinh(y)
-dt/sqrt[4 t^2 - 1] = -1/2 dy
The integral is thus y/2 + c =
arccosh(2t) + c =
arccosh(2/u) + c =
Log[2/u + sqrt(4/u^2 - 1)] + c =
Log[2 + sqrt(4 - u^2)] - Log(u) + c =
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