Another difficult integral/Also question on checking lower limit.

[Zula110100100]

Homework Statement

$\displaystyle{r^{2}\int^{\arcsin\frac{r}{t}}_{0^{+}}\frac{\sqrt{2}\sin(\theta-\frac{\pi}{4})\sin(2\arcsin(\frac{t}{r}\sin\theta))}{1-\cos(2\theta)}d\theta + \int^{\arcsin\frac{r}{t}}_{0^{+}}\frac{\cos(\frac{\pi}{4}-\theta)}{1-\cos(2\theta)}d\theta - \int^{\arcsin\frac{r}{t}}_{0^{+}}\arcsin(\frac{t}{r}\sin\theta)d\theta + \int^{\arcsin\frac{r}{t}}_{0^{+}}d\theta}$

$t = \sqrt{2}, r = 1$

This is once again beyond my skills for quite some time I imagine, but I am extremely curious to know the answer. The number empire integral calculator gives me that the second term approaches -∞ as theta approaches 0. Does this mean then that with 0+ I wind up with -∞ as the answer still? Or since the change in theta gets smaller too does it depend on the derivative? How do I find if there is a lower limit?

 

[lippyka]

Homework Equations \displaystyle{r^{2}\int^{\arcsin\frac{r}{t}}_{0^{+}}\frac{\sqrt{2}\sin(\theta-\frac{\pi}{4})\sin(2\arcsin(\frac{t}{r}\sin\theta))}{1-\cos(2\theta)}d\theta + \int^{\arcsin\frac{r}{t}}_{0^{+}}\frac{\cos(\frac{\pi}{4}-\theta)}{1-\cos(2\theta)}d\theta - \int^{\arcsin\frac{r}{t}}_{0^{+}}\arcsin(\frac{t}{r}\sin\theta)d\theta + \int^{\arcsin\frac{r}{t}}_{0^{+}}d\theta}The Attempt at a Solution I am not sure how to solve this problem.