#29 Measure you Calibre

$\large \displaystyle |x-1| \leq 3 \rightarrow \color{#20A900} \boxed{\large \displaystyle -1 \leq x \leq 3}$

$\large \displaystyle x^2 - y^2 = 1 \rightarrow \color{#20A900} \boxed{\large \displaystyle y = \pm \sqrt{x^2 - 1} }$

Which is a symmetrical curve with respect to the x-axis. So, the area we're looking for is:

$\large \displaystyle A = 2 \int_1^3 \sqrt{x^2-1} dx$

$\large \displaystyle \color{#20A900} x = \cosh(u) \rightarrow dx = \sinh(u) du$

$\large \displaystyle A = 2 \int_0^{acosh(3)} \sqrt{\cosh(u)^2-1} \sinh(u) du$

$\large \displaystyle A = 2 \int_0^{acosh(3)} \sinh^2(u) du$

$\large \displaystyle A = 2 \int_0^{acosh(3)} \frac{ \cosh(2u) - 1}{2} du$

$\color{#20A900} \boxed{ \large \displaystyle A = \frac{sinh(2u)}{2} \Bigg |_0^{acosh(3)} - acosh(3) }$

Now, to calculate $acosh(3)$ :

$\large \displaystyle u = acosh(3) \rightarrow \frac{e^u + e^{-u}}{2} = 3$

Set $\large \displaystyle e^u = v$

$\large \displaystyle v + \frac1v = 6 \rightarrow v^2 - 6v + 1 = 0 \rightarrow v = 3 + 2\sqrt{2} \rightarrow \color{#20A900} \boxed{ \large \displaystyle acosh(3) = \ln(|3 + 2\sqrt{2}|) }$

Also, if $\large \displaystyle e^u = v$ , $\large \displaystyle sinh(2u) = v^2 - \frac{1}{v^2} \rightarrow \color{#20A900} \boxed{ \large \displaystyle sinh(2u) = 12\sqrt{2}}$

So:

$\color{#20A900} \boxed{ \large \displaystyle A = 6\sqrt{2} + \ln(|3 + 2\sqrt{2}|) }$

Then:

$\large \displaystyle \color{#3D99F6} a = 6, b = 2, c = 3, d = 2, e = 2, \boxed{\large \displaystyle a+b+c+d+e = 15}$

We note that the minimum and maximum values of $|x-1|\le 2$ occur at $x_1=1$ and $x_2=3$ respectively. The area required is given by the integral below:

$\begin{aligned} I & = 2 \int_1^3 \sqrt{x^2-1} \ dx & \small \color{#3D99F6} \text{Let }x = \sec u, \ dx = \sec u \tan u \ du \\ & = 2 \int_0^\alpha \sec u \tan^2 u \ du & \small \color{#3D99F6} \text{where } \alpha = \sec^{-1} 3 \\ & = 2 \int_0^\alpha \sec u \left(\sec^2 u - 1\right) du \\ & = 2 {\color{#3D99F6} \int_0^\alpha \sec^3 u \ du} - 2 \int_0^\alpha \sec u \ du & \small \color{#3D99F6} \text{Using reduction formula (see footnotes)} \\ & = 2 {\color{#3D99F6} \left( \frac {\sin u \sec^2u}2 \bigg|_0^\alpha +\frac 12 \int_0^\alpha \sec u \ du \right)} - 2 \int_0^\alpha \sec u \ du \\ & = 6\sqrt 2 - \int_0^\alpha \sec u \ du \\ & = 6\sqrt 2 - \int_0^\alpha \frac {\sec u \tan u + \sec^2 u}{\tan u + \sec u} du \\ & = 6\sqrt 2 - \ln (\sec u + \tan u) \bigg|_0^\alpha \\ & = 6\sqrt 2 - \ln (3+2\sqrt 2) \end{aligned}$

$\implies A+B+C+D+E = 6+2+3+2+2 = \boxed{15}$

$$

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Reduction formula: $\small \displaystyle \int \sec^m x = \frac {\sin x \sec^{m-1}x}{m-1}+\frac {m-2}{m-1} \int \sec^{m-2} x \ dx$