Integration Grandmaster - P15

$\begin{aligned} A & = \int_1^2 \sqrt{x+\sqrt{x-1}}\ dx & \small \blue{\text{Let }u^2 = x = 1} \\ & = 2 \int_0^1 u\sqrt{u^2 + u + 1}\ du & \small \blue{\implies 2u \ du = dx} \\ & = 2 \int_0^1 u\sqrt{\left(u + \frac 12\right)^2 + \frac 34}\ du & \small \blue{\text{Let }u + \frac 12 = \frac {\sqrt 3}2\sinh t} \\ & = 2 \int_{\sinh^{-1}\frac 1{\sqrt 3}}^{\sinh^{-1}\sqrt 3} \left(\frac {\sqrt 3}2 \sinh t - \frac 12 \right) \cdot \frac 34 \cosh^2 t \ dt & \small \blue{\implies du = \frac {\sqrt 3}2 \cosh t\ dt} \\ & = \frac {3\sqrt 3}4 \blue{\int_{\sinh^{-1}\frac 1{\sqrt 3}}^{\sinh^{-1}\sqrt 3} \cosh^2 t \sinh t \ dt} - \frac 34 \int_{\sinh^{-1}\frac 1{\sqrt 3}}^{\sinh^{-1}\sqrt 3} \cosh^2 t \ dt & \small \blue{\text{Let }v = \cosh t} \\ & = \frac {3\sqrt 3}4 \blue{\int_{\frac 2{\sqrt 3}}^2 v^2 \ dv} - \frac 38 \int_{\sinh^{-1}\frac 1{\sqrt 3}}^{\sinh^{-1}\sqrt 3} (\cosh 2 t + 1) \ dt & \small \blue{\implies dv = \sinh t \ dt} \\ & = \left[\frac {\sqrt 3v^3}4 \right]_{\frac 2{\sqrt 3}}^2 - \left[\frac {3\sinh 2t}{16} + \frac {3t}8 \right]_{\sinh^{-1}\frac 1{\sqrt 3}}^{\sinh^{-1}\sqrt 3} \\ & = 2\sqrt 3 - \frac 23 - \frac {3\sqrt 3}4 + \frac 14 - \frac {3\sinh^{-1}\sqrt 3}8 + \frac {3\ln 3}{16} \\ & = \frac {60\sqrt 3-20+9\ln 3 - 18\sinh^{-1} \sqrt 3}{48} \approx 1.46052743 \end{aligned}$

Therefore $\lfloor 10000 A \rfloor = \boxed{14605}$ .

The integral is Taking $x-1=t^2$ $\int_1^2 \sqrt{x+\sqrt{x-1}}dx = \int_0^1 2t\sqrt{t^2 +t+1} dt = \int_0^1(2t+1)\sqrt{t^2+t+1} -\int_0^1 \sqrt{t^2+t+1}$ $=[ \frac{2}{3}{(t^2+t+1)}^{\frac{3}{2}} - \frac{2t+1}{4}\sqrt{t^2 +t+1} -\frac{3}{8} \sinh^{-1}{(\frac{2t+1}{\sqrt{3}})} ]_0^1$ $= 2\sqrt{3}-\frac{1}{3} -\frac{3\sqrt{3}-1}{4}-\frac{3}{8} (\sinh^{-1}(\sqrt{3}) -\sinh^{-1}{(\frac{1}{\sqrt{3}})})$ $= \boxed{1.4605}$ Answer is $\color{#20A900}[10000×1.4605]= 14605$ Note

$\int \sqrt{t^2 +t+1}dt = \frac{2t+1}{4} \sqrt{t^2+t+1} +\frac{3}{8} \sinh^{-1} (\frac{2t+1}{\sqrt{3}})$

$\begin{aligned} A=&\int_{1}^{2} \sqrt{x+\sqrt{x-1}} \, dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textrm{Let} \ \ u=\sqrt{x-1} \\ =&2\int_{0}^{1} u\sqrt{\left(u+\frac{1}{2}\right)^2+\frac{3}{4}} \, du \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textrm{Let} \ \ t=u+\frac{1}{2} \\ =& 2\int_{\frac{1}{2}}^{\frac{3}{2}} t\sqrt{t^2+\frac{3}{4}} \, dt-\int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{t^2+\frac{3}{4}} \,dt \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textrm{Let} \ \ t=\sqrt{\frac{3}{4}}\tan \theta \\ =& \left . \dfrac{2(t^2+\frac{3}{4})}{3}\right|_{\frac{1}{2}}^{\frac{3}{2}} -\frac{3}{4}\int_{\frac{\pi}{3}}^\frac{\pi}{6} \sec^3 \theta \, d\theta \\ =& \left . \dfrac{2(t^2+\frac{3}{4})}{3}\right|_{\frac{1}{2}}^{\frac{3}{2}}- \frac{3}{4}\left( \left . \frac{1}{2} \sec \theta \tan \theta +\frac{1}{2} \ln |\sec \theta +\tan \theta | \right)\right|_{\frac{\pi}{3}}^{\frac{\pi}{6}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int \sec^3 x \, dx= \frac{1}{2} \left(\sec x \tan x +\ln|\sec x+\tan x|\right) +C\\ \approx& 1.4605274\ldots \end{aligned}$

Suppose that $0 < a < 2^{\frac23}$ . If we put $u = \sqrt{x-a} + \tfrac12a^2$ , then $u^2 + \tfrac14a(4-a^3) \; = \; x + a^2\sqrt{x-a} \hspace{2cm} \frac{dx}{du} \; = \; 2u-a^2$ and hence $\begin{aligned} A(a) & = \; \int_1^2 \sqrt{x + a^2\sqrt{x-a}}\,dx \; = \; \int_{u(1)}^{u(2)}\sqrt{u^2 + \tfrac14a(4-a^3)}(2u - a^2)\,du \\ & = \; \Big[\tfrac23\big(u^2 + \tfrac14a(4-a^3)\big)^{\frac32}\Big]_{u(1)}^{u(2)} - a^2\int_1^2 \sqrt{u^2 + \tfrac14a(4-a^3)}\,du \\ & = \; \Big[\tfrac23\big(u^2 + \tfrac14a(4-a^3)\big)^{\frac32}\Big]_{u(1)}^{u(2)} - \tfrac14a^3(4-a^3)\int_{v(1)}^{v(2)} \cosh^2v\,dv \\ & = \; \Big[\tfrac23\big(u^2 + \tfrac14a(4-a^3)\big)^{\frac32}\Big]_{u(1)}^{u(2)} - \tfrac18a^3(4-a^3)\int_{v(1)}^{v(2)} (\cosh2v+1)\,dv \\ & = \; \Big[\tfrac23\big(u^2 + \tfrac14a(4-a^3)\big)^{\frac32}\Big]_{u(1)}^{u(2)} - \tfrac18a^3(4-a^3)\Big[\sinh v \cosh v + v \Big]_{v(1)}^{v(2)} \\ & = \; \Big[\tfrac23\big(u^2 + \tfrac14a(4-a^3)\big)^{\frac32} - \tfrac12a^2u\big(u^2 + \tfrac14a(4-a^3)\big)^{\frac12}\Big]_{u(1)}^{u(2)} - \tfrac18a^3(4-a^3)\Big[v\Big]_{v(1)}^{v(2)} \\ & = \; \left[\tfrac23\big(x + a^2\sqrt{x-a}\big)^{\frac32} - \tfrac12a^2(\sqrt{x-a} + \tfrac12a^2)\sqrt{x + a^2\sqrt{x-a}}\right]_1^2 - \tfrac18a^3(4-a^3)\left[\sinh^{-1}\left(\frac{2u}{\sqrt{a(4-a^3)}}\right)\right]_{u(1)}^{u(2)} \end{aligned}$ In particular $\begin{aligned} A(1) & = \; \left[ \tfrac23\big(x + \sqrt{x-1}\big)^{\frac32} - \tfrac12\big(\sqrt{x-1} + \tfrac12\big)\sqrt{x + \sqrt{x-1}}\right]_1^2 - \tfrac38\left[\sinh^{-1}\big(\tfrac{2u}{\sqrt{3}}\big)\right]_{\frac12}^{\frac32} \\ & = \; \tfrac54\sqrt{3} - \tfrac{5}{12} - \tfrac38\sin^{-1}\sqrt{3} + \tfrac38\sinh^{-1} \big(\tfrac{1}{\sqrt{3}}\big) \end{aligned}$ and so $\lfloor 10000 A(1)\rfloor = \boxed{14605}$ .

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