- #1
issacnewton
- 1,000
- 29
Hi
I was trying to solve this problem. The integral is
[tex]\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi} [/tex]
Now I came across the integral while doing one of the problems in "Introduction to
Electrodynamics" By David Griffiths (3ed) . The substitution of course, as suggested by
Karl Weierstrass, is
[tex]t=\tan(\frac{\phi}{2}) [/tex]
But the new limits of the integral are same at lower and upper point. I checked the solution
manual of the book and the author says that
[tex]\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi}=2\int_{0}^{\pi} \frac{d\phi}{1+a \cos \phi}[/tex]
I couldn't understand this step. Any insight will be appreciated.
I was trying to solve this problem. The integral is
[tex]\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi} [/tex]
Now I came across the integral while doing one of the problems in "Introduction to
Electrodynamics" By David Griffiths (3ed) . The substitution of course, as suggested by
Karl Weierstrass, is
[tex]t=\tan(\frac{\phi}{2}) [/tex]
But the new limits of the integral are same at lower and upper point. I checked the solution
manual of the book and the author says that
[tex]\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi}=2\int_{0}^{\pi} \frac{d\phi}{1+a \cos \phi}[/tex]
I couldn't understand this step. Any insight will be appreciated.