Evaluating $\int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}.$

Question

I want to evaluate the following integral ($\theta_0>0$)
\begin{equation*}
\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta
\end{equation*}
So I thought about $d(\cos\theta)=-\sin\theta d\theta$, then
\begin{equation*}
\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta = – \int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}
\end{equation*}
My question is about that step. Is that legit? I mean, that integral is doubly improper at both limits. How can I ensure that it is integrable?

Answer 1

After making the substitutions $\cos(\theta)= x$ and $\cos(\theta_0)=x_0$, we have for $0<\theta_0<\pi$ and $-1<x<x_0$

$$\int_{\theta_0}^\pi \sqrt{\frac{1-\cos(\theta)}{\cos(\theta_0)-\cos(\theta)}}\,d\theta=\int_{-1}^{x_0}\sqrt{\frac{1}{(x_0-x)(1+x)}}\,dx$$

The integrand has square root singularities at $-1$ and $x_0$, and the integral exists as both an improper Riemann integral and a Lebesgue integral.