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 =
To evaluate the integral, you can start by performing the trigonometric substitution. Let's substitute u = 2sinθ. Then, du = 2cosθ dθ and √(4-u^2) = 2cosθ.
The integral becomes:
∫ du / (u√(4-u^2))
= ∫ 2cosθ dθ / (2sinθ * 2cosθ)
= ∫ dθ / sinθ
= ∫ cscθ dθ
Now, to integrate cscθ, we can use the method of integration by parts.
Let's rewrite cscθ as 1/sinθ:
∫ (1/sinθ) dθ
Now, using the formula for integration by parts: ∫ u dv = uv - ∫ v du, we can choose u = 1 and dv = cscθ dθ:
∫ (1/sinθ) dθ
= ∫ u dv
= uv - ∫ v du
= θ - ∫ (dθ/ sinθ)
The integral of (dθ/ sinθ) can be evaluated by substituting v = cosθ and dv = -sinθ dθ:
= θ - ∫ (-sinθ / sinθ) dθ
= θ + ∫ dθ
= θ + C
Since we made the substitution u = 2sinθ, we need to express the final answer in terms of u:
= arcsin(u/2) + C
Therefore, the final answer to the integral ∫ du / (u√(4-u^2)) is arcsin(u/2) + C.
The answer to the integral of du/(u*sqrt(4-u^2)) is:
Log[2 + sqrt(4 - u^2)] - Log(u) + c, where c is the constant of integration.