Let $K\geq2$ be some constant. Then I would like to show that the following integral is finite: \begin{equation} \int_{0}^{\sin^{-1}(1/K)} \int_{\theta}^{2\cos^{-1}(1-K\sin(\theta)) - \theta} \frac{\sin ^{2}\left(\frac{\theta+\varphi}{2}\right)}{\sin \theta \sin \varphi} d \varphi d \theta. \end{equation}
What I have tried: I have tried to use the identity $$\frac{\sin ^{2}\left(\frac{\theta+\varphi}{2}\right)}{\sin \theta \sin \varphi} = \frac{\sin\left( \frac{\theta}{2} \right)\cos\left( \frac{\varphi}{2} \right)}{4\cos\left( \frac{\theta}{2} \right)\sin\left( \frac{\varphi}{2} \right)} + \frac{1}{2} + \frac{\cos\left( \frac{\theta}{2} \right)\sin\left( \frac{\varphi}{2} \right)}{4\sin\left( \frac{\theta}{2} \right)\cos\left( \frac{\varphi}{2} \right)}$$ to split up the integral as follows $$\int_{0}^{\theta_{0}} \int_{\theta}^{2\cos^{-1}(1-K\sin(\theta)) - \theta} \frac{\sin\left( \frac{\theta}{2} \right)\cos\left( \frac{\varphi}{2} \right)}{4\cos\left( \frac{\theta}{2} \right)\sin\left( \frac{\varphi}{2} \right)}d \varphi d \theta \\+ \int_{0}^{\theta_{0}} \int_{\theta}^{2\cos^{-1}(1-K\sin(\theta)) - \theta}\frac{\cos\left( \frac{\theta}{2} \right)\sin\left( \frac{\varphi}{2} \right)}{4\sin\left( \frac{\theta}{2} \right)\cos\left( \frac{\varphi}{2} \right)} d \varphi d \theta + \frac{\pi^2}{2}.$$ Then in both integrals I first work out the inner integral (so with respect to $\varphi$). Then the for remaining integrals with variable $\theta$ I try to argue with for example L'Hôpital's rule and Taylor expansions of trigonometric functions at $\theta = 0$ and at $\theta = \sin(1/K)$ that the integrands are equal to $0$ in the integration bounds. After that I argue that the integrands are continuous on the rest of the integration intervals and thus conclude that the integrands are bounded and thus also the integrals.
My question: The above method seems really quite complicated. I am wondering if there is an easier way to show that \begin{equation} \int_{0}^{\sin^{-1}(1/K)} \int_{\theta}^{2\cos^{-1}(1-K\sin(\theta)) - \theta} \frac{\sin ^{2}\left(\frac{\theta+\varphi}{2}\right)}{\sin \theta \sin \varphi} d \varphi d \theta \end{equation} is finite. For example with just a direct sequence of estimates from above or some other significantly less complicated method.
Many thanks in advance.