Integral Problem: Solve $\int\sqrt{\frac{1+t^{2}}{1-t^{2}}}\,dt$

[footmath]

hello.please solve this integral:
$ \int\sqrt{\frac{1+t^{2}}{1-t^{2}}}\,dt $

 

[Ray Vickson]

hello.please solve this integral:
$ \int\sqrt{\frac{1+t^{2}}{1-t^{2}}}\,dt $

It cannot be done in terms of so-called "elementary" functions (powers, roots, trig functions, inverse trigs, logs, exponentials, etc.). Have you heard of Elliptic functions?

RGV

 

[footmath]

I have heard the Elliptic function .
please explain to solve this integral.

 

[Char. Limit]

Let t=sin(u) and then dt = cos(u) du. Substitute those in for every t and dt you find. Some stuff should cancel out, and what you have left is very close to the definition of the elliptic function (of the second kind), given below.

E(\phi, m) = \int_0^\phi \sqrt{1 - m sin^2(\theta)} d\theta

You just need to pick the right value for m.

 

[footmath]

this problem at the beginning was: int_(sinx)^1/2 which transformed to $ A=\int\sqrt{1+\sin^{2}x}\,dx $ -\int_1/{1+\sin^{2}x} and then transformed to $ \int\sqrt{\frac{1+t^{2}}{1-t^{2}}}\,dt $

 

[Char. Limit]

The form you'll want it in is \int \sqrt{1 + sin(\theta)^2} d\theta. Then, setting m=-1, you'll be able to put it in terms of the Elliptic Integral of the Second Kind.

 

[footmath]

would you please explain the solution of elliptic integral

 

[Char. Limit]

I just did. In post 4, set m=-1 and see what integral you get. It's strikingly similar to the integral you're trying to solve.

 

[footmath]

Thank you but I can not solve this integral:$ A=\int\sqrt{1+\sin^{2}x}\,dx $
please explain about solution .