Changing the limits of integration, getting 0

[Mr Davis 97]

I have the integral $\displaystyle \int_0^{2 \pi} \frac{1-\cos x}{3+\cos x} ~ dx$. I want to make the tangent half-angle substitution $t = \tan (x/2)$ so that I can get a rational function. However, both limits of integration just become zero. This is the first case. In the second case, I notice that by symmetry the integral is equivalent to $\displaystyle 2 \int_0^{\pi} \frac{1-\cos x}{3+\cos x} ~ dx$. With this case the substitution works just fine and I am able to evaluate the integral.

My question is, what if I just didn't happen to catch that symmetry argument? Would I just not be able to use the half-angle substitution, since both bounds go to zero? Is noticing the symmetry really the only way around this?

 

[haushofer]

Maybe this helps:

https://math.stackexchange.com/questions/929380/what-is-wrong-with-the-following-u-substitution

 

[haushofer]

So, for which t one has arctan(t)=pi?

I'd say it lies outside of arctan's range.

 

[Fred Wright]

I want to make the tangent half-angle substitution t=tan(x/2)

I tried this substitution and found it intractable. For the integral,
$$I=2\int_ {0}^{\pi}\frac {1-cos(x)} {3+ cos(x)}dx$$
Let $t=sin(\frac {x} {2})$ with $dt = \frac {1}{2}t \sqrt {1-t^2}dx$ and the integral becomes,
$$I= 2\int_{0}^{1} \frac {tdt}{\sqrt {1-t^2}(1-\frac {t^2} {2})}$$
Make another substitution with $u=t^2$ and $du=2tdt$ to get,
$$I=\int_{0}^{1}\frac {du}{\sqrt {u}(1- \frac {u} {2}) \sqrt {1-u}}$$
I compare this with the hypergeometric integral
$$B(b,c-b)\;_ {2}F_1(a,b;c;z)=\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}dx$$
where B is the beta function. I get $a=1$, $b= \frac {1}{2}$, $c=1$, $z=\frac {1}{2}$, $B(\frac {1}{2},\frac {1}{2}) = \pi$, to find
$$I=\pi \;_ {2}F_1(1,\frac {1}{2};1;\frac {1}{2})$$
I suspect there is a formula for giving this hypergeometric function as an algebraic number but I don't have access to the literature quoted on the Wikipedia page for the hypergeometric function.
Peace,
Fred

 

[Mr Davis 97]

So, for which t one has arctan(t)=pi?

I'd say it lies outside of arctan's range.

Good point. I guess my question now is, if I have an integral $\int_a^b f(x) ~dx$, and I want to make the change of variables $u = g(x)$ or $x = h(u)$, what are the formal restrictions on $h$ and $g$?

 

[Demystifier]

Good point. I guess my question now is, if I have an integral $\int_a^b f(x) ~dx$, and I want to make the change of variables $u = g(x)$ or $x = h(u)$, what are the formal restrictions on $h$ and $g$?

The domain of $g(x)$ (or equivalently, the codomain of $h(u)$) must contain the interval $(a,b)$.