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Omniscience

  • 3 years ago

\[\int tanxdx\] integration by parts

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  1. Omniscience
    • 3 years ago
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    i integrate it twice; but i dont know why it failed

  2. irkiz
    • 3 years ago
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    separate them into tanx and dx

  3. Omniscience
    • 3 years ago
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    i did\[(tanx)(x) - \int\limits xsec^{2}dx \]

  4. hartnn
    • 3 years ago
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    integral sin x /cos x put t= cos x

  5. Omniscience
    • 3 years ago
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    i know..but i want to do it by parts

  6. mayankdevnani
    • 3 years ago
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    - ln |cos x| + C proof int tan x dx = int sinx/cos x set u=cosx then we find du=-sinx dx substitute du=-sin x, u=cos x int sinx/cos x= - int of (-1) sin x dx / cosx =- int of du/u Solve the integral = - ln |u| + C substitute back u=cos x = - ln |cos x| + C

  7. mayankdevnani
    • 3 years ago
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    ok @Omniscience

  8. Omniscience
    • 3 years ago
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    i know u-sub i want to do it by parts

  9. mayankdevnani
    • 3 years ago
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    http://wiki.answers.com/Q/Integration_by_parts_of_x_tanx

  10. mayankdevnani
    • 3 years ago
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    i think it will help you!!!

  11. Omniscience
    • 3 years ago
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    that's \(xtanx\)

  12. mayankdevnani
    • 3 years ago
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    oh!!! sorry

  13. irkiz
    • 3 years ago
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    maybe you could try converting tan x to sinx/cosx. then integrate sinxsecx by parts?

  14. Omniscience
    • 3 years ago
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    ok i will try it

  15. irkiz
    • 3 years ago
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    wait. is the ans ln|secx|?

  16. Omniscience
    • 3 years ago
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    -ln(cosx)

  17. irkiz
    • 3 years ago
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    looks the same lol

  18. irkiz
    • 3 years ago
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    sec x is (cosx)^-1 bring the -1 to the front

  19. irkiz
    • 3 years ago
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    yup thats the answer

  20. Omniscience
    • 3 years ago
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    ok thanks

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