How Can I Solve This Annoyingly Difficult Integral for Homework?

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SUMMARY

The integral of dx/(1-x^2)^(3/2) can be solved using the substitution x=sin(θ), which transforms the integral into ∫sec²(θ) dθ. The original attempt at solving the integral involved recognizing the derivative of inverse sine, but the approach was complicated by the presence of the cube in the denominator. Ultimately, the correct method involves using trigonometric identities and integral tables, specifically the formula for ∫du/(a²-u²)^(3/2), which simplifies the problem significantly.

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Homework Statement

integrate dx/(1-x^2)^(3/2)

The attempt at a solution

I first noticed that the derivative of inverse sin equals (1 x^2)^(-1/2) and that my equation is that derivative cubed. However, I had no idea what to do with the cube, so I dicarded that thought.

Next I tried to do a u substitution. obviously I can't make u equal 1-x^2 because that leaves me with an extra x in my du that won't go away. I was inspired by remembering some powers of trig integrations to try to make du equal the derivative of inverse sin, but that also failed as I had no inverse sin in the original equation to replace with u and last I checked, I can't integrate du^3.

I finally checked my tables of integrals and found where the integral of du/(a^2-u^2)^(3/2) equals u/a^2(a^2-u^2). This seems to give me my answer, but I can't figure out how to derive that formula.
 
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From [tex]I=\int\frac{dx}{(1-x^2)^{\frac{3}{2}}}[/tex] apply the substitution [tex]x=\sin\theta[/tex]. Hint: you should get [tex]I=\int\sec ^2 \theta\, d\theta[/tex].
 
You probably substituted wrongly to get du^3
 

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