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stottrupbailey

  • 3 years ago

integrate 1/sec^2xtanx

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  1. stottrupbailey
    • 3 years ago
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    \[\int\limits_{}^{}\frac{ 1 }{ \sec^2xtanx }dx\]

  2. itsmylife
    • 3 years ago
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    \[\int\limits_{?}^{?} \frac{ \cos x }{ \sec x \tan x }=\int\limits_{?}^{?} \frac{ 1 }{ \sec x \tan x }* \cos x\] you can use integration by parts now, and sec x tan x is differentiation of sec x

  3. stottrupbailey
    • 3 years ago
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    umm how did you get that first integral, sorry I'm confused

  4. stottrupbailey
    • 3 years ago
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    or would I just turn the sec^2x into (1-tan^2x)?

  5. stottrupbailey
    • 3 years ago
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    \[\int\limits_{}^{}cotxcos^2xdx\]

  6. stottrupbailey
    • 3 years ago
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    is that basically the same thing?

  7. stottrupbailey
    • 3 years ago
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    I hate calculus!!!

  8. stottrupbailey
    • 3 years ago
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    sorry just needed to vent

  9. stottrupbailey
    • 3 years ago
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    it just disappeared?

  10. stottrupbailey
    • 3 years ago
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    I'm starting to wonder if maybe I did the original problem wrong to even get to this scenario.... the original integral was \[\int\limits_{}^{}\frac{ 188 }{ x^2\sqrt{16x^2-81} }dx\]

  11. RadEn
    • 3 years ago
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    int 1/sec^2xtanx dx = int cos^2 x cosx/sinx dx = int (1-sin^2 x)cosx/sinx dx = int cosx/sinx dx - int sinxcosx dx = int cosx/sinx dx - int 1/2*sin2x dx case I : int cosx/sinx dx int by u-sub let u=sinx ---> du = cosx dx so, int cosx/sinx dx = int du/u = ln(u) = ln(sinx) case II : - int 1/2*sin2x dx = -(-1/4 cos2x) = 1/4 cos2x so, the total = ln(sinx) + 1/4 cos2x

  12. RadEn
    • 3 years ago
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    opppss. i forgot + c :)

  13. stottrupbailey
    • 3 years ago
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    from there I got this: \[47\int\limits_{}^{}\frac{ 1 }{ x^2\sqrt{x^2-(\frac{ 9 }{ 4 })^2} }\]

  14. stottrupbailey
    • 3 years ago
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    so then I did \[x=\frac{ 9 }{ 4 }\sec\]

  15. stottrupbailey
    • 3 years ago
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    that should be sec theta

  16. stottrupbailey
    • 3 years ago
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    and from there I got \[47\int\limits_{}^{}\frac{ 1 }{ (\frac{ 9 }{ 4 })^2\sec^2\theta(\frac{ 9 }{ 4 })\tan \theta }d \theta \]

  17. stottrupbailey
    • 3 years ago
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    somebody stop me if I'm screwing up lol

  18. RadEn
    • 3 years ago
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    i have done it above @stottrupbailey happy valentine's day , hahaha

  19. stottrupbailey
    • 3 years ago
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    so then I get \[47 (\frac{ 4 }{ 9 })^3\int\limits_{}^{}\frac{ 1 }{ \sec^2\theta \tan \theta }d \theta\]

  20. stottrupbailey
    • 3 years ago
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    And thus the current dilemma, the integral of sec^2tan

  21. stottrupbailey
    • 3 years ago
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    I'm confused by RadEn's explanation, can someone maybe explain what they did there? I got lost when he broke it up into subtracting two integrals...

  22. RadEn
    • 3 years ago
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    sorry, i didnt see the original problem.. i just see the first posting

  23. RadEn
    • 3 years ago
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    well, ur original problems is int ( 188/(x^2sqrt(16x^2-81)) dx in the frist step u have right let x = 9/4 secθ ---> x^2=9/16 sec^2 x dx = 9/4 secθtanθ dθ and for sqrt(16x^2-81), it can be sqrt(81sec^2 θ -81) = sqrt(81(sec^2 θ - 1) = 9sqrt(tan^2 θ) = 9tanθ so, ur integral can be : |dw:1360827737581:dw|

  24. RadEn
    • 3 years ago
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    from the equation : x = 9/4 secθ or secθ = 4x/9, reprsented in the figure : |dw:1360828428409:dw|

  25. RadEn
    • 3 years ago
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    thus, the result is = 752/9 sinθ + c = 752/9 * sqrt(16x^2 - 81)/(4x) + c

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