How can the limits of this integral be changed to make it easier to solve?

[issacnewton]

Hi

I was trying to solve this problem. The integral is

$$\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi}$$

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

$$t=\tan(\frac{\phi}{2})$$

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

$$\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi}=2\int_{0}^{\pi} \frac{d\phi}{1+a \cos \phi}$$

I couldn't understand this step. Any insight will be appreciated.

 

[issacnewton]

wouldn't $\cos(\pi-x)=-\cos(x)$ ?

But looking at my original integral limits, or seeing that the substitution gives the same upper and lower limit, how would I think in your direction. I mean , when I see that the substitution gives same limits, I feel like stuck there. How do I analyze the situation mathematically ?

 

[micromass]

Hi

I was trying to solve this problem. The integral is

$$\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi}$$

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

$$t=\tan(\frac{\phi}{2})$$

But the new limits of the integral are same at lower and upper point.

This is because the substitution is not allowed in this case. You'll need the substitution function, that is $$\tan(\phi/2)$$ to be continuously differentiable on $$[0,2\pi]$$. But it isn't continuous in pi. This is not a problem with indefinite integrals however...

I checked the solution
manual of the book and the author says that

$$\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi}=2\int_{0}^{\pi} \frac{d\phi}{1+a \cos \phi}$$

I couldn't understand this step. Any insight will be appreciated.

Do

[tex]\int_{0}^{2\pi} \frac{d\phi}{1+a \cos \phi}$$=\int_{0}^{\pi} \frac{d\phi}{1+a \cos \phi}+\int_{\pi}^{2\pi} \frac{d\phi}{1+a \cos \phi}$$<br /> <br /> Now, in the last integral, do the substituation $$t=2\pi-\phi$$...[/tex]

 

[Dick]

wouldn't $\cos(\pi-x)=-\cos(x)$ ?

But looking at my original integral limits, or seeing that the substitution gives the same upper and lower limit, how would I think in your direction. I mean , when I see that the substitution gives same limits, I feel like stuck there. How do I analyze the situation mathematically ?

Yeah, I spoke a little too fast. Try cos(2pi-x)=cos(x). Take the integral over [pi,2pi] and substitute 2pi-u=theta.

 

[issacnewton]

micromass, I TOTALLY forgot about the continuity. Doing too much physics does that
to the person...lol

Dick, thanks for the input.