Satanic Integral?

Question

Integral of:

dx/[(x^4)(sqrt(a^2-x^2))]

Just giving me a trig identity that I can work with to start off would be great. I've tried all sorts of stuff. Integrating by parts didn't result in anything good. There's probably some form of u-substitution also involved, but I don't see where this could be done.

Answer 1

Wow that is a nasty one. In my calculus book there is a section about trig substitution in integrals. Integrals containing sqrt(a^2-x^2) you can eliminate the sqrt by substituting x = a sin u so that dx = a cos u du. So what you have would become...

∫(a cos u) / (a sin u)^4 sqrt(a^2 - a^2 sin^2 u) du

∫(a cos u) / (a sin u)^4 (a) sqrt(1 - sin^2 u) du

∫cos u / (a sin u)^4 sqrt(1 - sin^2 u) du

1/a^4 ∫cos u / sin^4 u (cos u) du

1/a^4 ∫1 / sin^4 u du

You can take it from here. Hope that helps.

Answer 2

An obvious start is x = a*cos(z). Using that will require you to integrate sec^4(z), which you can do by writing sec^4(z) = sec²(z)[1 + tan²(z)] = sec²(z) + sec²(z) tan²(z). The first one integrates, the second one you do by parts (u = tan²(z), dv = sec²(z) dz) which gives you back sec^4(z) but with a negative sign, allowing you to solve for the integral of sec^4(z).

You should get

[-1/(3a²)]√(a² - x²) [(1/x³) + 2/(a²x)]

Good luck!

Answer 3

How bout:

a^2sin^2+a^2cos^2=a^2 ?

So

a^2-a^2 sin^2=a^2cos^2

or

a^2-a^2cos^2=a^2 sin^2

?