Question

Please help, I got stuck near the end of this trig substitution:

the integral of (1/(x^2*sqrt(x^2 - 9)))

Answer 1

\[\int\limits 1/x ^{2}\sqrt{x^{2}-9} dx\]

Answer 2

i did substitution:
\[\sec \theta=x/3\]
then \[\sqrt{(x/3)^{2}-1}=\tan (x)\]
try this one...

Answer 3

I came to this:
\[\int\limits d \Theta /9\sec \Theta\]

Answer 4

almost... I got:
\[1/3\int\limits_{ }^{}d \theta/\sec(\theta)=1/3\int\limits_{ }^{}\cos(\theta)d \theta\]

Answer 5

let's check coefficients:
dx=3 sec*tan
right?

Answer 6

yes

Answer 7

Good work. To use the traditional method, the one the instructor is drilling into you\[x =3\sec \theta\]

Answer 8

\[\int\limits_{ }^{}3\sec \theta*\tan \theta d \theta/9*\sec ^{2}\theta*\tan \theta=1/3\int\limits_{}^{}\cos \theta d \theta\]
did I lost something...?

Answer 9

I did set x =\[3\sec \theta\]

and I got \[9 \tan ^{2}\theta\] under the square root, which will become \[3 \tan \theta\]?

Answer 10

let's see:
of x=3 sec
then under integral we'll have:
\[dx=3 \sec \theta* \tan \theta d \theta\]
denominator:
\[9*\sec ^{2}\theta*\tan \theta\]
3/9=1/3
is it better now?

Answer 11

Sorry to interrupt what inik was doing. My way comes to\[(1/27)\int\limits_{}^{}1/(\sec ^{2}\theta \tan \theta)\]If you can somehow find a way to integrate that.

Answer 12

chaguanas, need to put proper dx in place as well - see above

Answer 13

inik, we have \[\sqrt{x ^{2}-9}\]

when replaced by 3sec , it becomes \[\sqrt{9\sec ^{2}-9}\]

Answer 14

ok
=\[3\sqrt{\sec ^{2} \theta-1}\]
\[=3\tan \theta\]

Answer 15

and we also have \[x ^{2}\] outside the square root which makes the whole denominator \[9 \sec ^{2}\theta 3 \tan \theta \]

Answer 16

oh yeah, thanks inik, I need to make some adjustments

Answer 17

Eventually it becomes\[(1/9)\int\limits_{}^{}d \theta/\sec \theta\]

Answer 18

\[x = 3\sec\theta\]\[dx = 3\sec\theta\tan\theta d\theta\]
\begin{eqnarray*}\int\frac{1}{x^2\sqrt{x^2-9}}dx &=& \int \frac{1}{(3\sec\theta)^2\sqrt{9\sec^2\theta - 9}}3\sec\theta\tan\theta d\theta\\&=&\int \frac{3\sec\theta\tan\theta}{9\sec^2\theta \cdot 3\tan\theta}d\theta\\&=&\int\frac{1}{9\sec\theta}d\theta\\&=&\frac{1}{9}\int\cos\theta d\theta\\&=&\frac{1}{9}\sin{\theta}+K\\&=&\frac{1}{9}\sin(\mathop{\mathrm{arcsec}}(x/3)))+K\\&=&\frac{1}{9}\sqrt{1-\frac{9}{x^2}}+K\\&=&\frac{\sqrt{x^2-9}}{9x}+K.\end{eqnarray*}

Answer 19

agreed!
I lost one 3... :(( sorry

Answer 20

thank you so much for going thru this long problem with me.

Answer 21

welcome and good job on your part!!