# Thread: Help with an integral

1. ## Help with an integral

Hi.

Can anybody help me with the solution of the next integral:

$\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi$

where $\theta=\theta(\phi)$.

The problem is that don't know the relation between $\theta$ and $\phi$.

2. Originally Posted by guest
Hi.

Can anybody help me with the solution of the next integral:

$\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi$

where $\theta=\theta(\phi)$.

The problem is that don't know the relation between $\theta$ and $\phi$.
looks somewhat like an arc length integral for a polar function ... in what context was this integral set up?

3. lets try this: call the value of the definate integral f, square it and evaluate it as a double integral. the square root of the double integral should be f also lets pretend that theta and phi are independant since we havent got a relationship between them

$f=\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi$
$f^2=(\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi)^2$
$f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta(\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2) (d \phi)^2$ $f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta\sin^2 \thetad\phi^2+\int_{\phi_0}^{\phi}\int_{\phi_0}^{\ phi}\cos^2\thetad\theta^2$
$f^2=\int_{\phi_0}^{\phi}(\phi-\phi_0)\cos^2\theta\sin^2\theta*d\phi+\int_{\phi_0 }^{\phi}\frac{2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0}{4}d\theta$
$f^2=(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}$
finally
$f= \sqrt{(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}}$

i cannot guarentee this is the answer youre looking for, however this is the only way i was able to integrat the function $e^{-x^2}$
good luck!!

4. Originally Posted by skeeter
looks somewhat like an arc length integral for a polar function ... in what context was this integral set up?
This is a part of the formula for the speed down an unknown path on a sphere with friction.

5. Originally Posted by JeffN12345
$f=\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi$
$f^2=(\int_{\phi_0}^{\phi} \cos\theta\sqrt{\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2} d \phi)^2$
$f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta(\sin^2 \theta+\left(\frac{d\theta}{d\phi}\right )^2) (d \phi)^2$ $f^2=\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi} \cos^2\theta\sin^2 \theta d\phi^2+\int_{\phi_0}^{\phi}\int_{\phi_0}^{\phi}\c os^2\theta d\theta^2$
$f^2=\int_{\phi_0}^{\phi}(\phi-\phi_0)\cos^2\theta\sin^2\theta*d\phi+\int_{\phi_0 }^{\phi}\frac{2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0}{4}d\theta$
$f^2=(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}$
finally
$f= \sqrt{(\frac{\phi^2-\phi_0^2}{2}-\phi_0)\cos^2\theta\sin^2\theta+\frac{(2(\phi-\phi_0)+\sin2\phi-\sin2\phi_0)(\phi-\phi_0)}{4}}$
Your intgrating would be okay if $\theta \neq \theta(\phi)$. I think that because of not knowing the relation $\theta = \theta(\phi)$ this integral can't be solved annalyticaly.