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Evaluate the integral (in comments)
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\[\int\limits \frac{ 1 }{ x^2 (x^2-4)^{1/2} }dx\]
I did a trig sub for x=2sec(theta)
\[\int\limits \frac{ 2\tan \theta \sec \theta }{ (2\sec \theta)^2\sqrt(4\sec^2 \theta -4) } d \theta\]
That is correct Just break down the expression into sine and cosine terms. It'll be greatly simplified.
So i get it down to \[\int\limits \frac{ 1 }{ 4\sec \theta } d \theta\] is that correct?
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That is absolutely correct.
\[\frac{ \sin \theta }{4 }\] is what i get, but how do i get it in terms of x
sec(theta) = x/2 cos(theta) = 2/x Use this : sin(theta) = sqrt(1- cos^2 (theta)) You can find value of sin(theta) in terms of x from here.
\[\frac{ \sqrt{x^2-4} }{ 4x }+C\]
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