Integral $\int_1 ^{h_1}\frac{1}{\sqrt{h}\sqrt{h^2-b}} dh$

Question

I am stuck at the flowing analytical integration: $\int_1 ^{h_1}\frac{1}{\sqrt{h}\sqrt{h^2-b}} dh$ where $0<b<1$ and $h_1>1$.

This can be recast as $C\int_1 ^{x_1} \frac{\sqrt{x}}{\left((1-b) x^3+ b\right)^{3/4}} dx$ if we assume $h=\sqrt{(1-b)x^3+b}$. C is some constant.

In Mathematica, I am getting Elliptic (upper one) or Hypergeometric ${}_2 F_1$ functions. I am not able to get those functions analytically by myself. I need your help to calculate those integrations (analytically). Mostly I need the answer for indefinite integration, then one can put the limits.

Thanks in advance.

Thanks. Can you kindly illustrate the steps which lead to the known elliptic integral? — Satadru Bag, Jun 15 '17 at 12:30

score 1 · Accepted Answer · answered Jun 16 '17 at 19:40

1

Just put $h=\sqrt b\sec\phi$ in the indefinite integral to get $$\int \frac{dh}{\sqrt h\sqrt {h^2-b}}=b^{-1/4}\int \sqrt{\sec\phi} d\phi$$ Next look here : Integral of root of $\sec x$

answered Jun 16 '17 at 19:40

Nikunj

6,160

Integral $\int_1 ^{h_1}\frac{1}{\sqrt{h}\sqrt{h^2-b}} dh$

1 Answers1