Find $\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg) \, \, \, dx \, \, dy$ Find $$\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg)dx \, \, dy$$
So obviously using cylindrical is the way to go to give $\theta r$ inside the integral (after considering the jacobian term).
But how can the limits of $x$ be $(x,1)$. I am guessing it is meant to say $dy \, \, dx$ right so those limits are for $y$? Please tell me if you disagree. 
But even if this is the case, it is not possible to compute the integral.
After sketching out $x<y<1$ and $0<x<1$, I realised that the region is the triangle bounded by origin, $(0,1)$ and $(1,0)$. So the limits of $r$ and $\theta$ are $(0,1/\cos\theta)$ and $(\pi/4 , \pi /2)$. So we have $$\int \limits_{\pi/4}^{\pi/2} \int \limits_0^{1/\cos\theta} r\theta \, \, \, dr \, \, d\theta$$ which gives $$\int \limits_{\pi/4}^{\pi/2} \frac 12 \theta \sec^2 \theta \, \, \,  d\theta$$
The integral of $\sec^2 \theta $ is $ln |\cos \theta |$ and one of the limits is $\pi/2$ which is undefined. Please help.

Let $I=\int_{0}^{1} \int_{x}^{1} arctan(\frac{y}{x}) dy dx  $, then considering polar coordinates we have the foolowing region : 
so as we see  from  the picture, the region is bounded by $\theta$ between  $\frac{\pi}{4} $ and  $\frac{\pi}{2} $. Now taking a ray from the origin crossing the region, we recignize that it starts from origin and finaly left the region meeting the line $y=1$, so that  $r\sin \theta= 1$, i.e. $ r $  is bounded  between $0$ and  $\frac{1}{\sin \theta}$. so 
$$I = \int_{\frac{\pi}{4} }^{\frac{\pi}{2}} \int_{ 0}^{\frac{1}{\sin \theta} } r \theta dr d\theta = \int_{\frac{\pi}{4} }^{\frac{\pi}{2}} \frac{\theta}{ 2 sin^2 \theta } d\theta= - \frac{\theta}{2\tan \theta } \mid_{\frac{\pi}{4}}^{\frac{\pi}{2}} + \int_{\frac{\pi}{4} }^{\frac{\pi}{2}}  \frac{d\theta}{2\tan \theta } = -\lim_{\theta \rightarrow \frac{\pi}{2} }   \frac{\theta}{2\tan \theta } + \frac{\pi}{8 }  + \frac{1}{2} \ln (\sin \theta)\mid_{\frac{\pi}{4}}^{\frac{\pi}{2}}=  \frac{\pi}{8 } -\frac{1}{2} \ln (\sin \frac{\pi}{4}). $$
