Calculus Problem

  • #1
iggybaseball
57
0
I am having trouble solving this integral:

[tex]\frac{\1}{(4-z^2)^(3/2)}[/tex]

I know that x = asin(theta)
theta = arcsin (x/a)
d(theta) = 1 / sqrt(4-z^2) dz

but then I get stuck. Could someone give me a hand?

ps there should be a number 1 on top of the fraction and the integral has dz after it respectfully. I couldn't get these two in ( It is the first time I used latex) Thanks
 

Answers and Replies

  • #2
George Jones
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Choose z = asin(theta). With regards to your problem, what should you choose a to be? Differentiate z = asin(theta) to find dz = ?

Regards,
George
 
  • #3
HallsofIvy
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It's the same thing but simpler: let z= 2sin(θ) then dz= 2cos(θ).
[tex]\frac{1}{(4-z^)^{\frac{3}{2}}}= \frac{1}{8(1- sin^2(\theta))^{\frac{3}{2}}}[/tex][tex]= \frac{1}{8 cos^3(\theta)}[/tex]
The integral becomes [tex]\frac{1}{8}\integral{\frac{d\theta}{cos^2(\theta)}[/tex].

The way you were doing it works too, of course.
Since [tex]d\theta= \frac{dz}{\sqrt{((4-z^2)}}[/tex] and [tex]\frac{1}{(4-z^2)^{\frac{3}{2}}}= \frac{1}{4-z^2}\frac{1}{\sqrt{4-z^2}}[/tex], that gives the same thing.
 
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