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farukk

  • 3 years ago

integrate tan^3 xsec x dx...!

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  1. farukk
    • 3 years ago
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    \[\int\limits \tan^3 x secx dx\]

  2. farukk
    • 3 years ago
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    what to do?

  3. experimentX
    • 3 years ago
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    \[ \tan^3 x \sec x = \tan x \sec x ( \sec^2 x - 1)\] seems like it will get the same

  4. farukk
    • 3 years ago
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    yeah ..next step

  5. Shadowys
    • 3 years ago
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    note that d(sec x) =sec x tan x dx

  6. farukk
    • 3 years ago
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    yes all right

  7. farukk
    • 3 years ago
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    what's next..

  8. Shadowys
    • 3 years ago
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    let u = sec x then proceed. du= sec x tan x dx so it all becomes \(\int (u^2 - 1) du\)

  9. farukk
    • 3 years ago
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    dear are you solving integral by parts?

  10. farukk
    • 3 years ago
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    or by substituion?

  11. farukk
    • 3 years ago
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    it will become like this i guess u^3/3 - u +c

  12. Shadowys
    • 3 years ago
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    \( \int (\sec^2-1) \tan x \sec x dx\) =\( \int (\sec^2-1) d(\sec x)\) =\( \int (u^2-1) du\), where u=\sec x\) by substitution.

  13. farukk
    • 3 years ago
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    yes next step..

  14. Shadowys
    • 3 years ago
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    eh? \(\int u^n du= \frac{u^{n+1}}{n+1} +C\)

  15. farukk
    • 3 years ago
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    after applying we ll get as i above wrote ?

  16. Shadowys
    • 3 years ago
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    you'll get the answer....from integrating the above eq.

  17. farukk
    • 3 years ago
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    wait let me write what i am getting.

  18. farukk
    • 3 years ago
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    \[\frac{ \sec^3x }{ 3 }- secx +c = answer ???\]

  19. Shadowys
    • 3 years ago
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    it should be, yeah. :) you can always check your answers with wolfram.

  20. farukk
    • 3 years ago
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    wolfram.?

  21. Shadowys
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=integrate%20tan%5E3%20x%20sec%20x%20dx&t=crmtb01

  22. farukk
    • 3 years ago
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    but answer is differnet in my book?

  23. farukk
    • 3 years ago
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    \[\frac{ 1 }{ 3 }(secxtan^2x-2secx)+c =answer ( book)\]

  24. farukk
    • 3 years ago
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    @Shadowys

  25. Shadowys
    • 3 years ago
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    just take sec x out. and also, the integral, \(\int du=u\)

  26. farukk
    • 3 years ago
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    did't get..

  27. Shadowys
    • 3 years ago
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    \(\int (u^2-1) du = \int u^2 du - \int du\) following the integral, it becomes, \(\frac{u^3}{3} - u+C\)

  28. farukk
    • 3 years ago
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    i have already written this.Scroll up a bit

  29. farukk
    • 3 years ago
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    i just wanna get the same answer as in my book...:)

  30. Shadowys
    • 3 years ago
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    oh. unfortunately, either your book's wrong, or the computer integration is wrong. because wolfram agrees with you....lol

  31. farukk
    • 3 years ago
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    yeah i have seen there but what you think about book's anwer is that wrong,there is really less possibilit though and yes lolz?

  32. farukk
    • 3 years ago
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    i have three books with same answer(:

  33. Shadowys
    • 3 years ago
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    lol i dun think wolfram's wrong, in anycase....lol

  34. farukk
    • 3 years ago
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    if we solve this by ''integration by parts'' rather tha by substituion?

  35. farukk
    • 3 years ago
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    \[\int\limits uvdx= u \int\limits vdx - \int\limits(\int\limits vdx) u'dx\]

  36. Shadowys
    • 3 years ago
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    same way, but this time i still let du=d sec x tan x dx. and v=sec^2 -1 sorry, gtg, but i think you'll still get the same answer...(or that's what my calc tells me)

  37. farukk
    • 3 years ago
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    @Shadowys okay i think my all books are wrong this time.(there is no benifit or edge having more than one books) :)!

  38. farukk
    • 3 years ago
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    thanks i took your time

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