How Can I Solve This Annoyingly Difficult Integral for Homework?

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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 I=\int\frac{dx}{(1-x^2)^{\frac{3}{2}}} apply the substitution x=\sin\theta. Hint: you should get I=\int\sec ^2 \theta\, d\theta.
 
You probably substituted wrongly to get du^3
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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