1. ## Integral Help

I have the following integral i cant find the solution.
I have tried differents ways to solve but nothing.

a is constant and it will take values on the interval [0,1]

2. Hi

$\displaystyle \sqrt{1+a^2\sin^2x}$ looks like $\displaystyle \sqrt{1+\sinh^2X}=\cosh X$. Hence, you may try substituting $\displaystyle a\cdot\sin x = \sinh u \Leftrightarrow u=\mathrm{arcsinh}(a\cdot\sin x)\Leftrightarrow x=\arcsin\left(\frac{\sinh u}{a}\right)$, it should give something easier to evaluate.

3. Originally Posted by marta_fdez
I have the following integral i cant find the solution.
I have tried differents ways to solve but nothing.

a is constant and it will take values on the interval [0,1]

The answer is given below, where $\displaystyle E[x | m]$ is an elliptic integral of the second kind.

4. Originally Posted by mr fantastic
The answer is given below, where $\displaystyle E[x | m]$ is an elliptic integral of the second kind.
I'm searching information about elliptic integrals of second kind. That would make sense as the integral arose when I was trying to calculate the arc length of a curve generated when a point is moving along the surface of a sphere, with determined polar and azimuthal angular speed (both constants).

flyingsquirrel I also tried that variable change at the beggining (i thought the same thing), but i didn't reach anything.

$\displaystyle \sqrt{1+a^2\sin^2x}$
We subsitute $\displaystyle u=\mathrm{arcsinh}(a\cdot\sin x)\Leftrightarrow x=\arcsin\left(\frac{\sinh u}{a}\right) \Rightarrow \mathrm{d}x=\frac{1}{a}\cdot\frac{\cosh u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}$
We get $\displaystyle \int \sqrt{1+a^2\sin^2x}\,\mathrm{d}x=\frac{1}{a}\int \sqrt{1+\sinh^2 u}\frac{\cosh u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}\,\mathrm{d}u=\frac{1}{a}\int \frac{\cosh^2 u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}\,\mathrm{d}u$
At this point, I realize that, when I did it earlier, I forgot a $\displaystyle \cosh u$... that's why it was easy to integrate. Sorry