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Math Help - [SOLVED] am I thinking right??

  1. #1
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    Exclamation [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?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by codex123 View Post
    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 \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 \int\sec t\,dt, you will need to apply this trick:

    \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 z=\sec t+\tan t, we end up with \int\frac{\,dz}{z}=\ln\!\left|z\right|+C\implies \int\sec t\,dt=\ln\!\left|\sec t+\tan t\right|+C

    So what do you think your final answer should be?
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  3. #3
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    Thanks!

    Yeah I skipped a step but that is where I messed up. Thanks for helping out! So the final answer would be
    ln|(x/3 + sqrt(x^2 -9)/3 | + C
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