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rishabh.mission Group Title

Find ?????

  • one year ago
  • one year ago

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  1. minecraftchick78 Group Title
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    what?

    • one year ago
  2. rishabh.mission Group Title
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    \[\int\limits_{ }^{}( \sqrt \cot + \sqrt tanx).dx\]

    • one year ago
  3. UnkleRhaukus Group Title
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    not a nice problem

    • one year ago
  4. rishabh.mission Group Title
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    ohh then plz solve

    • one year ago
  5. UnkleRhaukus Group Title
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    where did you get the question from?

    • one year ago
  6. rishabh.mission Group Title
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    in Sample Paper

    • one year ago
  7. UnkleRhaukus Group Title
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    that dosent tell me anything

    • one year ago
  8. UnkleRhaukus Group Title
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    can you do it again @Mimi_x3 ?

    • one year ago
  9. Mimi_x3 Group Title
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    \[\int \sqrt{tanx}+\sqrt{cotx}dx =>\int\frac{sinx+cosx}{\sqrt{sinxcosx}} => \int\frac{sinx+cosx}{\sqrt{sin2x}}\] \(u=sinx-cosx\) =>\(du=(sinx+cosx)dx\) \(1-sin2x =u^2\) => \(sin2x = (1-u^2)\) Therefore, \[ \int\frac{1}{\sqrt{1-u^2}}du\]

    • one year ago
  10. amoodarya Group Title
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    arcsin(sinx-cosx) +const

    • one year ago
  11. Mimi_x3 Group Title
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    i have a feeling that something is wrong :|

    • one year ago
  12. youarestupid Group Title
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    I would help if i could, sorry

    • one year ago
  13. amoodarya Group Title
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    arcsin(sinx-cosx) +const this is not answer i write it for you to test "85 Mimi_x3" answer but its not true

    • one year ago
  14. amoodarya Group Title
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    this integral has no answer in simple way

    • one year ago
  15. amoodarya Group Title
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    i turn it to complex form but its not easy

    • one year ago
  16. amoodarya Group Title
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    you diffrenciate this sqrt2 arcsin(sinx-cosx) i think its correct

    • one year ago
  17. Mimi_x3 Group Title
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    Yeah, i made a minor error; sorry! Happens, when i dont do it on paper lol \(sin2x = 2sinxcosx\) \(sinxcosx = 1/2sin2x \) \[\int\limits\frac{\sin+cosx}{\sqrt{1/2\sin2x}} =>\sqrt{2} \int\limits\frac{sinx+cosx}{\sqrt{\sin2x}} =>\sqrt{2}\int\limits\frac{1}{\sqrt{1-u^{2}}} du\]

    • one year ago
  18. rishabh.mission Group Title
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    OK but can i solve this in a different method like can i Re write as \[\int\limits_{}^{} \sqrt{ \tan x} (1+\cot x )dx\] and then Put tan x = t², so that sec²x dx = 2t dt Or \[ dx=2t . dt / 1+t ^{4} \]

    • one year ago
  19. shubhamsrg Group Title
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    NCERT example if I am recalling correctly eh ?

    • one year ago
  20. Tushara Group Title
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    why did u re write it in that form?... it dusnt look right.......... btw, thats a good question, a challenging one

    • one year ago
  21. rishabh.mission Group Title
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    yeah it isz

    • one year ago
  22. Mimi_x3 Group Title
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    @rishabh.mission: the solution i gave; is the easiest method to solve this integral.

    • one year ago
  23. rishabh.mission Group Title
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    ok thnku

    • one year ago
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