# Question 5b52a

Dec 15, 2017

$I = \setminus {\int}_{3}^{4} \frac{\mathrm{dx}}{\setminus} \sqrt{{x}^{2} - 9} = \setminus {\tan}^{- 1} \left(\setminus \sqrt{\frac{7}{2}}\right) .$

#### Explanation:

I = \int_3^4(dx)/\sqrt{x^2-9}; \qquad\qquad t \equiv \sqrt{x^2-9}; \qquad dx = (tdt)/\sqrt{9+t^2};

I = \int_0^\sqrt{7} (cancel{t}dt)/(cancel{t}\sqrt{9+t^2}); \qquad\qquad t \equiv 3\tanhu; \qquad dt = 3\sech^2u.du#
$\setminus \sqrt{9 + {t}^{2}} = \setminus \sqrt{9 + 9 \setminus {\tanh}^{2} u} = 3 \setminus \sech u$

$I = \setminus {\int}_{0}^{\setminus {\tanh}^{- 1} \left(\setminus \frac{\sqrt{7}}{3}\right)} \frac{\cancel{3} \setminus {\sech}^{\cancel{2}} u}{\cancel{3} \cancel{\setminus \sech u}} . \mathrm{du}$

$I = \setminus {\int}_{0}^{\setminus {\tanh}^{- 1} \left(\setminus \frac{\sqrt{7}}{3}\right)} \setminus \sech u . \mathrm{du} = {\left[\setminus {\tan}^{- 1} \left(\setminus \sinh u\right)\right]}_{0}^{\setminus {\tanh}^{- 1} \left(\setminus \frac{\sqrt{7}}{3}\right)}$

$\setminus \sinh u = \setminus \tanh \frac{u}{\setminus} \sqrt{1 - \setminus {\tanh}^{2} u}$

$\setminus \quad = {\left\{\setminus {\tan}^{- 1} \left(\setminus \tanh \frac{u}{\setminus} \sqrt{1 - \setminus {\tanh}^{2} u}\right)\right\}}_{0}^{\setminus {\tanh}^{- 1} \left(\setminus \frac{\sqrt{7}}{3}\right)}$
$\setminus \quad = \setminus {\tan}^{- 1} \left(\frac{\setminus \frac{\sqrt{7}}{3}}{\setminus \sqrt{1 - {\left(\setminus \frac{\sqrt{7}}{3}\right)}^{2}}}\right) - \setminus {\tan}^{- 1} \left(0\right) = \setminus {\tan}^{- 1} \left(\setminus \sqrt{\frac{7}{2}}\right)$