How to find a useful variable change for this integral I would like to find the area of the following region
$$
D=\left \{(x,y): -\sqrt{1+y^2}\leq x\leq \sqrt{1+y^2}; -1\leq y\leq (x+1)/2\right \}.
$$
I try to calculate the double integral brute force, but following this path
I came across some very unpleasant integrals. So I think that maybe a variable
change may be an appropriate approach here.
Can someone suggest me a useful variable change to calculate the area of this region?


See the diagram. You need to find the area of the region ABCO. If you integrate along $y$ axis taking strips of thickness $dy$ parallel to $x$ axis, you can see that from $y = -1$  to $y = 0$, both the left and right ends are bound by the hyperbola but for strips at $0 \leq y  \leq \frac{4}{3}$, the left is bound by the line $2y = x+1$ and the right is bound by the hyperbola. So we divide our integral in two parts.
Now to find the intersection of the line $2y = x+1$ and the hyperbola $x^2 - y^2 = 1$,
$x^2 - y^2 = 1$
At intersection, $x^2 - \frac{(x + 1)^2}{4} = 1$. That gives us $x = \frac{5}{3}, y = \frac{4}{3}$
So, $A = \displaystyle \int_{-1}^0 ({x_r - x_l}) \, dy \, \, + \int_{0}^{4/3} ({x_r - x_l}) \, dy$
$A = \displaystyle \int_{-1}^0 (\sqrt{1+y^2} - (-\sqrt{1+y^2}) \, dy \, \, + \int_{0}^{4/3} (\sqrt{1+y^2} - (2y-1)) \, dy$
$A = \displaystyle 2\int_{-1}^0 \sqrt{1+y^2} \, dy \, \, + \int_{0}^{4/3} (\sqrt{1+y^2} - 2y + 1) \, dy$
To integrate $\sqrt{1+y^2}$, one of the ways is to substitute $y = \tan \theta$.
Integral of $\sqrt{1+y^2}$ is given by  $\frac{y}{2} \sqrt{1+y^2} + \frac{1}{2} \ln ({y + \sqrt{1+y^2}})$.
You can check WolframAlpha for the same.
