# Calculus

posted by
**Tracy** on
.

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 just 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 =