Question

Having trouble with this one, particularly on after the trig substitution. What the heck are the bounds on the new integrand supposed to be? o_O
\[\huge \int\limits_{3\sqrt{2}}^{6} \frac{1}{t^3\sqrt{t^2-9}}\]

Answer 1

using a secant subsitutions, so dt = sec\(\theta\) tan\(\theta\) d\(\theta\)

Answer 2

i.e.: t=6 so sec\(\theta\) = 6 which... something with \(\pi\)

Answer 3

Is this correct so far?
\[\huge\int\limits_{3\sqrt{2} \rightarrow ?}^{6 \rightarrow ?} \frac{\sec \theta \tan \theta}{\sec^3 \theta \tan \theta} d \theta\]

Answer 4

let's see, the correct sub you want it\[t=3\sec u\implies dt=3\sec u\tan udu\]

Answer 5

is*

Answer 6

so your bounds are\[3\sqrt2=3\sec u\implies u=\cos^{-1}(\frac{\sqrt2}2)=\frac\pi4\]\[6=3\sec u\implies u=\cos^{-1}(\frac12)=\frac\pi3\]

Answer 7

so the integral is\[\int_{3\sqrt2}^6{dt\over t^3\sqrt{t^2-9}}=\int_{\pi/4}^{\pi/3}{3\sec u\tan udu\over3^3\sec^3u\sqrt{9\sec^2 u-9}}\]

Answer 8

ack so t = sec\(\theta\) was an error...

if
t = 3sec\(\theta\)
then
t = 6 \(\rightarrow\) cos\(\theta\) = 1/(6/3) = 1/2

Answer 9

yep ")

Answer 10

Awesome, thank you! :-D I can take it from here.

Answer 11

great :D
welcome!