# [SOLVED] am I thinking right??

• Feb 22nd 2009, 04:27 PM
codex123
[SOLVED] am I thinking right??
The integral 1/(sqrt((x^2)-9)

so far I have this....

x = 3sec(t)
dx = 3sec(t)tan(t)dt
this gives me --> integral (3sec(t)tan(t))dt/9(tan^2)(t)
1/3 integral sec(t)dt/tan(t) --> 1/3 integral csc(t)dt
1/3(-csc(t)cot(t)) + C = 1/3( -x/(sqrt(x^2 - 9)) * 3/(sqrt(x^2 - 9))

Please any help would be much appreciated!!! I don't know if I messed up or something. Any ideas?
• Feb 22nd 2009, 04:36 PM
Chris L T521
Quote:

Originally Posted by codex123
The integral 1/(sqrt((x^2)-9)

so far I have this....

x = 3sec(t)
dx = 3sec(t)tan(t)dt
this gives me --> integral (3sec(t)tan(t))dt/9(tan^2)(t)
1/3 integral sec(t)dt/tan(t) --> 1/3 integral csc(t)dt
1/3(-csc(t)cot(t)) + C = 1/3( -x/(sqrt(x^2 - 9)) * 3/(sqrt(x^2 - 9))

Please any help would be much appreciated!!! I don't know if I messed up or something. Any ideas?

Everything is fine up to the part in red. Your integral should be $\displaystyle \int\frac{3\sec t\tan t\,dt}{\sqrt{9\tan^2t}}=\int\frac{3\sec t\tan t\,dt}{3\tan t}=\int \sec t\,dt$

Now, to integrate $\displaystyle \int\sec t\,dt$, you will need to apply this trick:

$\displaystyle \int\sec t\,dt=\int\sec t\frac{\sec t+\tan t}{\sec t+\tan t}\,dt=\int\frac{\sec^2t+\sec t\tan t}{\sec t+\tan t}\,dt$. Applying the substitution $\displaystyle z=\sec t+\tan t$, we end up with $\displaystyle \int\frac{\,dz}{z}=\ln\!\left|z\right|+C\implies \int\sec t\,dt=\ln\!\left|\sec t+\tan t\right|+C$

• Feb 22nd 2009, 05:34 PM
codex123
Thanks!
Yeah I skipped a step but that is where I messed up. Thanks for helping out! (Happy) So the final answer would be
ln|(x/3 + sqrt(x^2 -9)/3 | + C