# Thread: Hard Trig Sub Need Help asap

1. ## Hard Trig Sub Need Help asap

Integral of 1/[(x^5)(9x^2-1)^(.5)] dx from sqrt(2)/3 to 2/3.
everyway i try it, it seems to be wrong!! i have been doing this problem for hours! i need help

here is my work: http://i198.photobucket.com/albums/a...n/scan0002.jpg
http://i198.photobucket.com/albums/a...scan0003-1.jpg

also I am not sure if #15 on the second link is correct either.

2. You forgot to change the limits of integration after you made the substitution.

$81 \int_{ \pi /4}^{\pi /3} \cos^{4} \theta \ d \theta$

3. ## ?

wait how did you do that?
it got me the right answer, but how did you change the limits?

4. Originally Posted by Jasonium
wait how did you do that?
$\frac {\sqrt{2}}{3} = \frac {1}{3} \sec \theta$

$\sqrt{2} = \sec \theta$

$\frac {1}{\sqrt{2}} = \cos \theta$

$\theta = \frac {\pi}{4}$

$\frac {2}{3} = \frac {1}{3} \sec \theta$

$\frac {1}{2} = \cos \theta$

$\theta = \frac {\pi}{3}$

5. Originally Posted by Jasonium
Integral of 1/[(x^5)(9x^2-1)^(.5)] dx from sqrt(2)/3 to 2/3.
everyway i try it, it seems to be wrong!! i have been doing this problem for hours! i need help

here is my work: http://i198.photobucket.com/albums/a...n/scan0002.jpg
http://i198.photobucket.com/albums/a...scan0003-1.jpg

also I am not sure if #15 on the second link is correct either.
First substitute u = 3x with dx = (1/3) du

so the integrand becomes
$
\frac{3^5}{u^5\sqrt{u^2-1}} \frac{1}{3} du$

Then substitute:

$u = \sec \theta$

note that:
$du = \sec \theta \tan \theta$

and the integrand becomes:
$
\frac{3^4}{(\sec \theta)^5 \tan \theta} \sec \theta \tan \theta$

which reduces to

$\frac{81}{(\sec \theta)^4}
$

OR
$
81 (\cos \theta)^4$

The integral of the fourth power of the cosine function is:

(12 x + 8 Sin[2 x] + Sin[4 x])
----------------------------
32

I hope this helps.