Problem:
Change the order of integration of
$$\int_0^{\pi/2}\int_0^{\cos(\theta)}\ \cos{(\theta)}\ dr\,d\theta$$
Solution:
First, I've made a plot of the given region:
$$0\leqslant\ r \leqslant \cos(\theta)$$
$$0\leqslant\ \theta \leqslant \pi/2$$
I have tried to define the new limits,
$$\int_0^1\int_0^{\cos(\theta)}\ \cos(\theta)\ d\theta\, dr$$
Some suggestions, tips,... to understand how to define limits when angles and trigonometric functions are involved in the original limits?
You need to get clear about your picture: the horizontal axis is the $\theta$-axis; that's where you need to put $\theta=0$ and $\theta=\pi/2$. The vertical axis is the $r$-axis, the curve should only go as high as $r=1$, not as high as $\pi/2 \approx 1.57.$
For any fixed value of $\theta$, the value of $r$ goes from $0$ up to $\cos\theta$. That's a vertical line segment in your picture.
For any fixed value of $r$, the value of $\theta$ goes from $0$ up to $\arccos r$. That's a horizontal line in your picture.
So you get $\displaystyle\int_0^1 \left( \int_0^{\arccos r} \cos\theta \,d\theta \right) \,dr.$
Your only mistake is using $\cos(\theta)$ in the bounds.
When we integrate with respect to $r$ first, the bounds are
$r=0$ and $r=\cos(\theta)$
However, when we integrate with respect to $\theta$ first, the bounds are
$\theta = 0$ and $\theta = \cos^{-1}(r)$ or $\theta = \arccos(r)$.
