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# Thread: Definite integral on elliptic integral where modulus is function of variable

1. How to prove:

$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}}$

where $k(\theta)=\sqrt\frac{4Rh \tan \theta}{Z^2+(R+h \tan \theta)^2}$

and $K[k(\theta)]$ is the complete elliptic integral of the first kind, defined by

$K[k(\theta)]= \int_0^{\frac{\pi }{2}}\frac{\,d\phi}{\sqrt{1-k^2(\theta)\sin^2 \phi}}$

and h, R and Z $\gt 0$

2. Can you show us what you have tried and where you are stuck? This will give our helpers a better idea how to provide help without perhaps offering suggestions that you may already be trying.

Originally Posted by bshoor
How to prove:

$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}}$

where $k(\theta)=\sqrt\frac{4Rh \tan \theta}{Z^2+(R+h \tan \theta)^2}$

and $K[k(\theta)]$ is the complete elliptic integral of the first kind, defined by

$K[k(\theta)]= \int_0^{\frac{\pi }{2}}\frac{\,d\phi}{\sqrt{1-k^2(\theta)\sin^2 \phi}}$

and h, R and Z $\gt 0$
Please make correction of the post:
$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}d\theta=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}}$

Originally Posted by MarkFL
Can you show us what you have tried and where you are stuck? This will give our helpers a better idea how to provide help without perhaps offering suggestions that you may already be trying.
I have tried in different ways. But the integral becomes more and more complicated. Anyway the identity can be modified to :

$\int_{0}^{\frac{\pi }{2}}{\sqrt {\sin \theta cos\theta}k(\theta) K[k(\theta)]}d\theta=\pi \sqrt{ \frac{Rh}{R^2 + (h+Z)^2}}$

by multiplying both side by $2 \sqrt{Rh}$

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