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anonymous

  • one year ago

integral 1/(1-tan^2x)

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  1. anonymous
    • one year ago
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    \[\int\limits_{}^{}1/(1-\tan^2x)\]

  2. anonymous
    • one year ago
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    I think we need to do u = 1-tan^2x I'm trying that now

  3. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{1-\tan^2{x}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x}{\cos^2x}-\dfrac{\sin^2x}{\cos^2{x}}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x-\sin^2x}{\cos^2{x}}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1\times \cos^2x}{\dfrac{\cos^2x-\sin^2x}{\cos^2{x}}\times \cos^2x}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{\cos^2x-\sin^2{x}}~dx}\)

  4. SolomonZelman
    • one year ago
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    no it is a mess my brain is not working right now

  5. anonymous
    • one year ago
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    It's ok this problem is also making my brain hurt lol

  6. Empty
    • one year ago
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    Hmmm is it being the difference of two squares helpful at all?

  7. anonymous
    • one year ago
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    How about cos2x = cos^2x - sin^2x And cos^2x= cos2x+sin^2x

  8. SolomonZelman
    • one year ago
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    Empty i thought of that, but what would u then do

  9. anonymous
    • one year ago
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    \[\int\limits_{}^{}(\cos2x + \sin^2x)/\cos2x\]

  10. anonymous
    • one year ago
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    Integral of 1 + integral of sin^2x/cos2x

  11. SolomonZelman
    • one year ago
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    cos²x ?

  12. SolomonZelman
    • one year ago
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    or cos(2x) ?

  13. anonymous
    • one year ago
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    I'm confused which cos you're talking about

  14. SolomonZelman
    • one year ago
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    both

  15. anonymous
    • one year ago
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    cos2x = cos^2x - sin^2x I used it to change cos^2x/(cos^2x - sin^2x)

  16. anonymous
    • one year ago
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    I can't see what to do if we do use the difference of squares

  17. Empty
    • one year ago
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    that was a glitch it double posted idk why lol

  18. SolomonZelman
    • one year ago
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    yeah difference of squares doesn't get me anywhere, then what sub would I make? and go into world's worse du=....

  19. anonymous
    • one year ago
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    I used wolfram to solve it and based on the answer there should be a logarithm using u substitution

  20. anonymous
    • one year ago
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    And a 1 somewhere

  21. anonymous
    • one year ago
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    oh wait. If at the beginning, we do u = tanx Then its du = sec^2xdx du=(1+tan^2x)dx So it would be \[\int\limits_{}^{}1/((1-u^2)(1+u^2))\]

  22. anonymous
    • one year ago
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    And that would be solvable with partial fractions I think

  23. SolomonZelman
    • one year ago
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    no master

  24. SolomonZelman
    • one year ago
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    sorry to dissapoint you, but your dx = (will be replaced by) cos²(idk what)

  25. SolomonZelman
    • one year ago
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    lol, du=sec^2x dx cos^2x du=dx cos^2(tan\(^{-1}u\)) du=dx

  26. anonymous
    • one year ago
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    u = tanx du=sec^2xdx cos^2xdu=dx So \[\int\limits_{}^{}\cos^2x/(1-u^2)du\]

  27. SolomonZelman
    • one year ago
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    can not have that-:( x's with u' and du like that

  28. anonymous
    • one year ago
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    But can't we just put the cos^2x as sec^2x in the denominator. And then sec^2x = 1+tan^2x = 1+u^2

  29. anonymous
    • one year ago
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    Even if you substitute it back? I didn't know

  30. SolomonZelman
    • one year ago
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    u= tan(x) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{1-u^2}~du}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+\tan^2x)(1-u^2)}~du}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+u^2)(1-u^2)}~du}\)

  31. SolomonZelman
    • one year ago
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    yes, sounds like it works.... no partial fractions is the intitive.

  32. SolomonZelman
    • one year ago
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    good job, i should say, master. indeed

  33. SolomonZelman
    • one year ago
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    now, partial fractions*

  34. anonymous
    • one year ago
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    Thank you lol

  35. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+\tan^2x)(1-u^2)}~du}\) \(\large\color{slate}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\frac{2}{u^2+1}+\frac{1}{u+1}-\frac{1}{u-1}du}\)

  36. SolomonZelman
    • one year ago
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    oh, the tan^2x is u^2

  37. SolomonZelman
    • one year ago
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    so if that substitution actually works as we said, the rest should be easy.

  38. anonymous
    • one year ago
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    Yup that matches what wolfram said the answer was.

  39. anonymous
    • one year ago
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    Good job and thank you :)

  40. SolomonZelman
    • one year ago
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    Lol, I need to get back to one variable integration... this is a good exercise.

  41. SolomonZelman
    • one year ago
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    I had another idea, that I messaged to Empti

  42. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{1}{1-x} =\sum_{n=0}^{\infty}x^n }\) \(\large\color{black}{ \displaystyle \frac{1}{1-\tan(x)} =\sum_{n=0}^{\infty}\tan^n(x) }\) \(\large\color{black}{ \displaystyle \frac{1}{1-\tan^2(x)} =\sum_{n=0}^{\infty}\tan^{2n}(x) }\) \(\large\color{black}{ \displaystyle \int\frac{1}{1-\tan^2(x)} dx=\int \sum_{n=0}^{\infty}\tan^{2n}(x)dx }\)

  43. SolomonZelman
    • one year ago
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    oh, no then, this becomes a difficult integral on the right

  44. SolomonZelman
    • one year ago
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    oh, you will learn it soon then.... it is just a power series representation thing very easy. good luck

  45. anonymous
    • one year ago
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    Thanks and to you as well

  46. SolomonZelman
    • one year ago
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    TH\(\color{blue}{\unicode{x13D49}}\)NK \(\color{blue}{\unicode{x3D4}}\)\(\color{blue}{O}\)U

  47. SolomonZelman
    • one year ago
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    cu

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