Radical Integral Of Radicals

$\int_3^6 \left ( \sqrt{x + \sqrt{12x-36}} + \sqrt{x - \sqrt{12x-36}} \ \right ) \ dx$

put $t=x-3$ . Rewrite given integral in terms of $t$ .

$\Rightarrow \int _{ 0 }^{ 3 } \left( \sqrt { t+3+2\sqrt { 3t } } +\sqrt { t+3-2\sqrt { 3t } } \right) dt$

$=\int _{ 0 }^{ 3 } \left( \sqrt { { (\sqrt { t } +\sqrt { 3 } ) }^{ 2 } } +\sqrt { { (\sqrt { 3 } -\sqrt { t } ) }^{ 2 } } \right) dt\\ =\int _{ 0 }^{ 3 } \left( \sqrt { t } +\sqrt { 3 } +\sqrt { 3 } -\sqrt { t } \right) dt \\ (\because \sqrt { { (\sqrt { 3 } -\sqrt { t } ) }^{ 2 } } =\sqrt { 3 } -\sqrt { t } \quad \forall \quad t\in (0,3))\\ =\int _{ 0 }^{ 3 } 2\sqrt { 3 } dt=6\sqrt { 3 } \\ \Rightarrow a\times b=\boxed { 18 }$

let $f(x) = \sqrt{x+\sqrt{12x-36}} + \sqrt{x-\sqrt{12x-36}}$ ,

we have

$\begin{aligned} f(x)^2 &=x+\sqrt{12x-36} + 2\sqrt{(x+\sqrt{12x-36})(x-\sqrt{12x-36})} + x - \sqrt{12x-36} \\&= 2x + 2\sqrt{x^2-(12x-36)} \\&= 2x + 2\sqrt{(x-6)^2} \\&= 2x + 2(6-x)\quad\quad (\sqrt{a^2}=a \space\space \forall a \geq 0) \\&= 12 \end{aligned}$

thus

$\displaystyle \int_3^6 f(x)\mathrm dx = (6-3)\sqrt{12} = \boxed{6\sqrt{3}}$

We can also use $u^2 = x - 3$ , then:

$\begin{cases} x = 6 & \Rightarrow u = \sqrt{3} \\ x = 3 & \Rightarrow u = 0 \\ x = u^2 + 3 & \Rightarrow dx = 2u \space du \end{cases}$

Let the integral be $I$ , then:

$\begin{aligned} I & = \int_3^6 \left(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}} \right) dx \\ & = \int_0^{\sqrt{3}} \left(\sqrt{u^2 + 3 + \sqrt{12}u} +\sqrt{u^2+3-\sqrt{12}u} \right) 2u \space du \\ & = \int_0^{\sqrt{3}} \left(\sqrt{u^2 + 2\sqrt{3}u+3} +\sqrt{u^2-2\sqrt{3}u+3} \right) 2u \space du \\ & = \int_0^{\sqrt{3}} \left(\sqrt{(\sqrt{3}+u)^2} +\sqrt{(\sqrt{3}-u)^2} \right) 2u \space du \\ & = \int_0^{\sqrt{3}} \left(\sqrt{3}+u + \sqrt{3}-u \right) 2u \space du \\ & = \int_0^{\sqrt{3}} 4 \sqrt{3} u \space du \\ & = 2\sqrt{3}\left[u^2\right]^{\sqrt{3}_0} \\ & = 6\sqrt{3} \end{aligned}$

$\Rightarrow ab = 6\times 3 = \boxed{18}$ .

We have $I=\displaystyle\int_{3}^{6}\left(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}\right)\,dx$ Puting $12x-36 =u^2 \Rightarrow x = \frac{u^2+36}{12}$ and differentiating we get $\,dx =\frac{1}{12}u\,du$ Rewriting the expressions $I=\displaystyle\int_{3}^{6}\left(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}\right)\,dx$ $I=\displaystyle\int_{3}^{6}\left(\sqrt{\frac{u^2+36}{12}+u}+\sqrt{\frac{u^2+36}{12}-u}\right)\frac{1}{12} u \,du$ $I=\frac{1}{ 12}\displaystyle\int_{3}^{6}\left(\sqrt{\frac{(u+6)^2}{12}}+\sqrt{\frac{(u-6)^2}{12}}\right)u\,du$

$I=\frac{1}{12}\displaystyle\int_{3}^{6}\left(\sqrt{\frac{(u+6)^2}{12}}+\sqrt{\frac{(u-6)^2}{12}}\right)u\,du$ $I = \frac{1}{12\sqrt 12}\displaystyle\int_{3}^{6} 2u^2\,du$ $I = \frac{2}{3\times 12\sqrt 12}\left[u^3\right]_{3}^{6}$

$I = \frac{2}{3\times 12\sqrt 12}\left[12x-3\sqrt{12x-3}\right]_{3}^{6}$ $I = \frac{36\times6}{6\sqrt 12}\Rightarrow 6\sqrt 3 = a\sqrt b$

Therefore, $ab = 6\times3 \implies 18$

Let the function we are integrating be K. Then, K^2 = 2x + 2(x^2-12x+36)^0.5. Since our values of x range from 3 to 6, we want the negative square root, so K^2 = 2x - 2(x-6) = 12. So K = 2 (3^.5). So our antiderivative is 2 (3^.5)x, giving us an answer of 2 (3^.5)(6-3) = 6 (3^.5). Therefore, 6*3 = 18.

The title of the problem made it very easy maybe it should be changed.