# How do you find the integral of ( 3x^2 - 18x + 20 )^(1/2)?

Sep 22, 2015

$I = \frac{7}{2 \sqrt{3}} \left(- \sin m {\sec}^{2} m + \ln | \frac{1 + \tan \left(\frac{m}{2}\right)}{1 - \tan \left(\frac{m}{2}\right)} |\right) + C$

$m = a r c \sec \left(\frac{\sqrt{3} \left(x - 3\right)}{\sqrt{7}}\right)$

#### Explanation:

$3 {x}^{2} - 18 x + 20 = 3 {x}^{2} - 18 x + 27 - 7 = 3 {\left(x - 3\right)}^{2} - 7 =$
$= {\left(\sqrt{3} \left(x - 3\right)\right)}^{2} - 7 = 7 \left({\left(\frac{\sqrt{3} \left(x - 3\right)}{\sqrt{7}}\right)}^{2} - 1\right)$

So, we have:

$I = \int \sqrt{7 \left({\left(\frac{\sqrt{3} \left(x - 3\right)}{\sqrt{7}}\right)}^{2} - 1\right)} \mathrm{dx}$

$I = \sqrt{7} \int \sqrt{{\left(\frac{\sqrt{3} \left(x - 3\right)}{\sqrt{7}}\right)}^{2} - 1} \mathrm{dx}$

$t = \frac{\sqrt{3} \left(x - 3\right)}{\sqrt{7}} \implies \mathrm{dt} = \frac{\sqrt{3} \mathrm{dx}}{\sqrt{7}} \implies \mathrm{dx} = \sqrt{\frac{7}{3}} \mathrm{dt}$

$I = \frac{7}{\sqrt{3}} \int \sqrt{{t}^{2} - 1} \mathrm{dt}$

$t = \sec m \implies \mathrm{dt} = \sec m \tan m \mathrm{dm}$
${t}^{2} - 1 = {\sec}^{2} m - 1 = {\tan}^{2} m$

$I = \frac{7}{\sqrt{3}} \int \sqrt{{\tan}^{2} m} \sec m \tan m \mathrm{dm}$

$I = \frac{7}{\sqrt{3}} \int {\tan}^{2} m \sec m \mathrm{dm} = \frac{7}{\sqrt{3}} \int {\sin}^{2} \frac{m}{\cos} ^ 3 m \mathrm{dm}$

$u = \sin m \implies \mathrm{du} = \cos m \mathrm{dm}$

$\mathrm{dv} = \sin \frac{m}{\cos} ^ 3 m \mathrm{dm} \implies v = \int \frac{\mathrm{dk}}{k} ^ 3 = - \frac{1}{2 {\cos}^{2} m}$

$I = \frac{7}{\sqrt{3}} \left(- \sin \frac{m}{2 {\cos}^{2} m} + \frac{1}{2} \int \frac{\cos m \mathrm{dm}}{\cos} ^ 2 m\right)$

$I = \frac{7}{\sqrt{3}} \left(- \sin \frac{m}{2 {\cos}^{2} m} + \frac{1}{2} \int \frac{\mathrm{dm}}{\cos} m\right)$

$\tan \left(\frac{m}{2}\right) = p \implies m = 2 \arctan p \implies \mathrm{dm} = \frac{2}{1 + {p}^{2}} \mathrm{dp}$

${p}^{2} = \frac{1 - \cos m}{1 + \cos m} = \frac{2}{1 + \cos m} - 1$
$1 + \cos m = \frac{2}{1 + {p}^{2}} \implies \cos m = \frac{1 - {p}^{2}}{1 + {p}^{2}}$

${I}_{1} = \int \frac{\mathrm{dm}}{\cos} m = \int \frac{\frac{2}{1 + {p}^{2}}}{\frac{1 - {p}^{2}}{1 + {p}^{2}}} \mathrm{dp}$
${I}_{1} = \int \frac{2}{1 - {p}^{2}} \mathrm{dp} = \int \frac{2}{\left(1 - p\right) \left(1 + p\right)} \mathrm{dp}$

$\frac{A}{1 - p} + \frac{B}{1 + p} = \frac{A + A p + B - B p}{\left(1 - p\right) \left(1 + p\right)}$

$A + B = 2 , A - B = 0 \iff A = 1 , B = 1$

${I}_{1} = \int \frac{\mathrm{dp}}{1 - p} + \int \frac{\mathrm{dp}}{1 + p} = - \ln | 1 - p | + \ln | 1 + p |$

${I}_{1} = \ln | \frac{1 + p}{1 - p} | = \ln | \frac{1 + \tan \left(\frac{m}{2}\right)}{1 - \tan \left(\frac{m}{2}\right)} |$

$I = \frac{7}{2 \sqrt{3}} \left(- \sin m {\sec}^{2} m + \ln | \frac{1 + \tan \left(\frac{m}{2}\right)}{1 - \tan \left(\frac{m}{2}\right)} |\right) + C$

$m = a r c \sec \left(\frac{\sqrt{3} \left(x - 3\right)}{\sqrt{7}}\right)$