Question

find integral of \frac{4}{3} \int\limits_{3}^{5}\sqrt{x^2 - 9}

Answer 1

\[\frac{4}{3} \int\limits\limits_{3}^{5}\sqrt{x^2 - 9} \]

Answer 2

Have you done trig substitutions? http://www.integralcalc.com/wp-content/uploads/2012/10/sin-tan-sec.png

Answer 3

it is hyperbolic sub. but im done well knowledgeable about how to execute the integral form there

Answer 4

not well knowledgeable* from*

Answer 5

Ah okay. I haven't done those in a while, but http://mathworld.wolfram.com/HyperbolicSubstitution.html and http://tutorial.math.lamar.edu/Classes/CalcI/DiffHyperTrigFcns_files/eq0007MP.gif

so let x = 3coshu

Answer 6

\[\large x = 3\cosh u\]
\[\large 1 = 3\sinh u \frac{du}{dx}\] \[\large dx = 3 \sinh u du\]

\[\large \int\limits \sqrt{x^2 - 9} dx = \int\limits\limits\limits \sqrt{9 \cosh^2 u - 9} ( 3 \sinh u) du\] (ignoring all the fractions and limits)

Answer 7

\[\large \int\limits\limits\limits\limits \sqrt{9( \cosh^2 u - 1)} ( 3 \sinh u) du \]
\[\large 9\int\limits \sqrt{\sinh^2 u} ( \sinh u) du = 9 \int\limits \sinh^2 u du \] Now you can prob integrate by changing the sinhu back to it's exponential form http://tutorial.math.lamar.edu/Classes/CalcI/DiffHyperTrigFcns_files/eq0003MP.gif