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?

split the area into 5 separate areas:
$$\int_{\pi}^{\pi+\tan^{-1}(1/\sqrt{2})}\frac{1}{2}r^2d\theta+
\int_{\pi+\tan^{-1}(1/\sqrt{2}))}^{-\tan^{-1}(1/\sqrt{2}))}\frac{1}{2}r^2d\theta+
\int_{-\tan^{-1}(1/\sqrt{2}))}^{0}\frac{1}{2}r^2d\theta+
\int_{0}^{\tan^{-1}(4/5)}\frac{1}{2}r^2d\theta+
\int_{\tan^{-1}(4/5)}^{\pi}\frac{1}{2}r^2d\theta
$$
the second and the last integral are triangles, and the first and the third are equal:
$$
\sqrt{2}+
2\int_{-\tan^{-1}(1/\sqrt{2}))}^{0}\frac{1}{2}r^2d\theta+
\int_{0}^{\tan^{-1}(4/5)}\frac{1}{2}r^2d\theta+
\frac{2}{3}
$$
the polar function that defines the hyperbola is:
$$r=\left(\frac{1}{\cos2\theta}\right)^{\frac{1}{2}}$$
so putting it into the integral then evaluating:
$$
\sqrt{2}+
2\int_{-\tan^{-1}(1/\sqrt{2}))}^{0}\frac{1}{2}\left(\frac{1}{\cos2\theta}\right)d\theta+
\int_{0}^{\tan^{-1}(4/5)}\frac{1}{2}\left(\frac{1}{\cos2\theta}\right)d\theta+
\frac{2}{3}
$$
$$
=\sqrt{2}+
\frac{1}{2}\ln \left(3\right)-\frac{1}{2}\ln \left(3-2\sqrt{2}\right)+
\frac{2}{3}
$$
