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 .