Question

Question #5b52a

Answer 1

`I = \int_3^4 (dx)/\sqrt{x^2-9}=\tan^{-1}(\sqrt{7/2}).`

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`
`\sqrt{9+t^2}=\sqrt{9+9\tanh^2u}=3\sechu`

`I = \int_{0}^{\tanh^{-1}(\sqrt{7}/3)}(cancel{3}\sech^cancel{2}u)/(cancel{3}cancel{\sechu}}.du`

`I = \int_0^{\tanh^{-1}(\sqrt{7}/3)}\sechu.du = [\tan^{-1}(\sinhu)]_0^{\tanh^{-1}(\sqrt{7}/3)}`

`\sinhu = \tanhu/\sqrt{1-\tanh^2u}`

`\quad = {\tan^{-1}(\tanhu/\sqrt{1-\tanh^2u})}_0^{\tanh^{-1}(\sqrt{7}/3)}`
`\quad = \tan^{-1}((\sqrt{7}/3)/(\sqrt{1-(\sqrt{7}/3)^2})) - \tan^{-1}(0) = \tan^{-1}(\sqrt{7/2})`