Integral Help

**marta_fdez** · May 7th 2008, 11:26 PM

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]

Thanks in advance.

**flyingsquirrel** · May 7th 2008, 11:59 PM

Hi

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

**mr fantastic** · May 8th 2008, 12:31 AM

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]

Thanks in advance.

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

**marta_fdez** · May 8th 2008, 01:40 AM

Originally Posted by mr fantastic

The answer is given below, where $E[x | m]$ 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.

Thanks for your answers.

**flyingsquirrel** · May 8th 2008, 02:55 AM

Originally Posted by marta_fdez

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

$\sqrt{1+a^2\sin^2x}$

We subsitute $u=\mathrm{arcsinh}(a\cdot\sin x)\Leftrightarrow x=\arcsin\left(\frac{\sinh u}{a}\right)<br /> \Rightarrow \mathrm{d}x=\frac{1}{a}\cdot\frac{\cosh u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}$

We get $\int \sqrt{1+a^2\sin^2x}\,\mathrm{d}x=\frac{1}{a}\int <br /> \sqrt{1+\sinh^2 u}\frac{\cosh u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}\,\mathrm{d}u=\frac{1}{a}\int <br /> \frac{\cosh^2 u}{\sqrt{1-\frac{\sinh^2 u}{a^2}}}\,\mathrm{d}u$

At this point, I realize that, when I did it earlier, I forgot a $\cosh u$ ... that's why it was easy to integrate. Sorry

Thread: Integral Help

LinkBack

Thread Tools

Display

Integral Help

Similar Math Help Forum Discussions

Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

Graph the region of integration and evaluate the double integral (integral from 0 to

Explaining result: comparing line integral and two-form integral

write [integral of] dx/SQRT(sinx) in terms of an elliptic integral

Evaluating a double integral to find its signle integral.

Search Tags