Question

integrate 3sqrt2 to 6 1/(x^3sqrt(x^2-9))dx

Answer 1

try \(=x^2\) and \(du=2xdx\) as a start

Answer 2

should i not use the trig sub x=3secø?

Answer 3

oh damn sorry nmv i didn't see it was a fraction
tilt

Answer 4

yes, you should
i make a mistake

Answer 5

where do i go after i sub to (3secøtanødø)/(3secøtan^2ø)

Answer 6

\[\frac{ 3\sec \theta \tan \theta d \theta \ }{ 3\sec \theta \tan^2 \theta }\]

Answer 7

problem started as \[\int\limits_{3\sqrt{2}}^{6} \frac{ 1 }{ x^3\sqrt{x^2-9} }dx\]

from which i subbed \[x=3\sec \theta\] \[dx = 3(\sec \theta \tan \theta) d \theta\] to get to \[\frac{ 3\sec \theta \tan \theta d \theta }{ 3\sec^3 \theta \tan^2 \theta}\]

Answer 8

try again

Answer 9

27 sec ^3 theta *

Answer 10

\[x^3\sqrt{x^2-9}\]

\[=(3\sec(\theta))^3\sqrt{(3\sec(\theta))^2-9}\]

\[=27\sec^3(\theta)\sqrt{9(\sec^2(\theta)-1)}\]
\[=81\sec^3(\theta)\sqrt{\tan^2(\theta)}\]
\[=81\sec^3(\theta)\tan(\theta)\]

Answer 11

ok thats right I see where I messed up but then where do I go with the problem from there?

Answer 12

simplify

Answer 13

then integrate

Answer 14

i can take the 3/81 out which leaves just \[du=\sec \theta \tan \theta\] in the numerator if my \[u=\sec \theta\] and that means my denom is \[u^2du\] then there are 2 du or an extra tangent?

Answer 15

\[\frac{3\sec(\theta)\tan(\theta)}{81\sec^3(\theta)\tan(\theta)}d\theta\]

\[=\frac{1}{27\sec^2(\theta)}d\theta=\frac{1}{2}\cos^2(\theta)d\theta\]

Answer 16

\[=\frac{1}{27\sec^2(\theta)}d\theta=\frac{1}{27}\cos^2(\theta)d\theta\]

Answer 17

wow, i totally got caught up in the trig to not see that reduce... thank you very much for your patience zarkon

Answer 18

you know what to do from here?

Answer 19

yeah