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?
 A: 
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.
A: 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:
$$
\sqrt{2}+
2\int_{-\tan^{-1}(1/\sqrt{2}))}^{0}\frac{1}{2}\left(\frac{1}{\cos2\theta}\right)d\theta+
\int_{0}^{\tan^{-1}(4/5)}\frac{1}{2}\left(\frac{1}{\cos2\theta}\right)d\theta+
\frac{2}{3}
$$
$$
=\sqrt{2}+
\frac{1}{2}\ln \left(3\right)-\frac{1}{2}\ln \left(3-2\sqrt{2}\right)+
\frac{2}{3}
$$
A: You don't need a double integral. A single integral is enough to calculate an area.
First, draw a picture. Having done that, split the problem into two by cutting the region along the $x$-axis and integrating along $y$:
$$A = \int_{-1}^0 x_{H'}-x_H\ dy + \int_0^k x_{H'}-x_L\ dy$$
where, for a given $y$, $x_H$ and $x_{H'}$ correspond to the left and right points on the hyperbola (i.e., $\mp\sqrt{1+y^2}$), and $x_L$ to the point on the line. The value $k$ is the $y$-co-ordinate of the point of intersection of the hyperbola and line.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&D \equiv
\braces{\!\!\pars{x, y} \mid -\root{1 + y^{2}} \!\leq\! x\! \leq\! \root{1 + y^{2}};\ -1\! \leq\! y\! \leq\! {x + 1 \over 2}}.
\\[2mm] &\ \bbox[#ffd,10px,border:1px groove navy]
{{\cal A}_{D} \equiv D\ area}:\ {\Large ?}.
\\ &
\end{align}

\begin{align}
{\cal A}_{D} & \equiv
\bbox[5px,#ffd]{\iint_{\mathbb{R}^{2}}
\bracks{\verts{x} \leq \root{1 + y^{2}}}
\bracks{-1 \leq y \leq {x + 1 \over 2}}\dd x\,\dd y}
\\[5mm] & =
\int_{-1}^{\infty}
\int_{-\root{\vphantom{A^{A}}1\ +\ y^{2}\,}}
^{\root{\vphantom{A^{A}}1\ +\ y^{2}\,}}
\bracks{x \geq 2y - 1}\dd x\,\dd y
\\[5mm] & =
\int_{-1}^{\infty}\ \overbrace{\bracks{2y - 1 < 
-\root{\vphantom{A^{A}}1\ +\ y^{2}\,}}}
^{\ds{= \bracks{y < 0}}}\
\int_{-\root{\vphantom{A^{A}}1\ +\ y^{2}\,}}
^{\root{\vphantom{A^{A}}1\ +\ y^{2}\,}}
\dd x\,\dd y
\\[2mm] & +
\int_{-1}^{\infty}\
\overbrace{\bracks{-\root{\vphantom{A^{A}}1\ +\ y^{2}\,} < 2y - 1 < 
\root{\vphantom{A^{A}}1\ +\ y^{2}\,}}}
^{\ds{= \bracks{0 < y < {4 \over 3}}}}\ \times
\\ & \phantom{\int_{-1}^{\infty}\ }
\int_{2y - 1}
^{\root{\vphantom{A^{A}}1\ +\ y^{2}\,}}\dd x\,\dd y
\\[5mm] & =
\underbrace{\int_{-1}^{0}2\root{1 + y^{2}}\dd y}
_{\ds{\root{2} + \on{arcsinh}\pars{1}}}\ +\
\underbrace{\int_{0}^{4/3}\bracks{\root{1 + y^{2}} - 2y + 1}\dd y}
_{\ds{{2 \over 3} + {\ln\pars{3} \over 2}}}
\\[5mm] & =
\bbx{{2 \over 3} + \root{2} + {1 \over 2}\ln\pars{3} +
\on{arcsinh}\pars{1}} \approx 3.5116 \\ &\
\end{align}

Note that
\begin{align}
\on{arcsinh}\pars{1} & = \ln\pars{1 + \root{2}} =
{1 \over 2}\ln\pars{3 + 2\root{2}}
\\[2mm] & =
-{1 \over 2}\ln\pars{3 - 2\root{2}}
\end{align}
