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RedPrince Group Title

what is the result of this question: sqrt{1-t^4}.2tdt

  • 2 years ago
  • 2 years ago

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  1. RedPrince Group Title
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    \[\int\limits \sqrt{1-t^4}.2tdt\]

    • 2 years ago
  2. .Sam. Group Title
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    \[u=t^2~~~and ~~~d u=2 td t\]

    • 2 years ago
  3. apoorvk Group Title
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    Use substitution: Let t^2 = m then 2tdt = dm The new integral is; \[\int\sqrt{1-m^2}dm\]

    • 2 years ago
  4. .Sam. Group Title
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    \[\int\limits \sqrt{1-u^2} \, du\] u=sin(a) \[\sqrt{1-u^2}\text{ = }\sqrt{1-\sin ^2(s)}\text{ = }\cos (s)\text{ and }s=\sin ^{-1}(u)\]

    • 2 years ago
  5. .Sam. Group Title
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    \[\int\limits \cos ^2(s) \, ds\] half angle \[\int\limits \left(\frac{1}{2} \cos (2 s)+\frac{1}{2}\right) \, ds\] separate \[\int\limits \frac{1}{2} \, ds+\frac{1}{2}\int\limits \cos (2 s) \, ds\] solve

    • 2 years ago
  6. RedPrince Group Title
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    you are right but I obtained it from: \[\int\limits \cos^2x \sqrt{sinx} dx\]

    • 2 years ago
  7. .Sam. Group Title
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    ?

    • 2 years ago
  8. RedPrince Group Title
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    @.Sam. don't understand how I get it!

    • 2 years ago
  9. RedPrince Group Title
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    @shivam_bhalla help me dear!

    • 2 years ago
  10. .Sam. Group Title
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    i didn't get \[\int\limits\limits \cos^2x \sqrt{sinx} dx\]

    • 2 years ago
  11. shivam_bhalla Group Title
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    @RedPrince , follow what @.Sam. told

    • 2 years ago
  12. RedPrince Group Title
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    @.Sam. integrate it! I get the result \[\int\limits \sqrt{1-t^4}.2tdt\] from \[∫\cos^2x \sqrt{sinx}dx\] by substituting \[sinx=t^2\]

    • 2 years ago
  13. RedPrince Group Title
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    @shivam_bhalla I also follow. you also?

    • 2 years ago
  14. .Sam. Group Title
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    its a long step to integrate this \[∫\cos^2x \sqrt{sinx}dx\]

    • 2 years ago
  15. shivam_bhalla Group Title
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    I simply remember that \[\int\limits_{}^{}\sqrt{a^2-x^2} = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}{(x/a)}\]

    • 2 years ago
  16. RedPrince Group Title
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    @.Sam. you guide me from where I ask question! see my uestion

    • 2 years ago
  17. shivam_bhalla Group Title
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    @RedPrince , see the above integral I wrote, substitute and simplify

    • 2 years ago
  18. RedPrince Group Title
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    @shivam_bhalla this is good and now the question is solved.....!! woooowwwwWWw. Thanksss!

    • 2 years ago
  19. .Sam. Group Title
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    you could use this too

    • 2 years ago
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  20. shivam_bhalla Group Title
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    @.Sam. , was just writing the same :)

    • 2 years ago
  21. .Sam. Group Title
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    or

    • 2 years ago
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  22. RedPrince Group Title
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    @.sam yes you are right. this is also helpful.

    • 2 years ago
  23. .Sam. Group Title
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    but that's just the formula without steps

    • 2 years ago
  24. shivam_bhalla Group Title
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    @RedPrince , I suggest you remember these formulas because they take a long time to derive. This is just formulas to quickly solve problems

    • 2 years ago
  25. shivam_bhalla Group Title
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    But , however I suggest you try to derive these formulas on your own first and then remember them

    • 2 years ago
  26. RedPrince Group Title
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    @.Sam. it doesn't matter. I will prove them myself. again thanks!

    • 2 years ago
  27. RedPrince Group Title
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    @shivam_bhalla yes, I will remember these formulas Inshallah.

    • 2 years ago
  28. asnaseer Group Title
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    RedPrince: I /think/ you may have made a mistake in your substitutions. I get this after the substitutions:\[\int\limits \sqrt{1-t^4}.2t^2dt\]

    • 2 years ago
  29. asnaseer Group Title
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    \[\sin(x)=t^2\]therefore:\[\cos(x)dx=2tdt\]

    • 2 years ago
  30. asnaseer Group Title
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    \[\cos^2(x)=1-\sin^2(x)=1-t^4\]

    • 2 years ago
  31. asnaseer Group Title
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    therefore:\[\cos^2(x)\sqrt{sin(x)}dx=\cos^2(x)t*\frac{2tdt}{cos(x)}=\cos(x)2t^2dt\]\[\qquad=\sqrt{1-t^4}2t^2dt\]

    • 2 years ago
  32. shivam_bhalla Group Title
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    @asnaseer , hawk eye :)

    • 2 years ago
  33. asnaseer Group Title
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    :)

    • 2 years ago
  34. shivam_bhalla Group Title
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    So our new integral should be \[m=t^2\] \[dm = 2tdt\] \[\int\limits_{}^{}\sqrt{1-m^2}\sqrt{m}dm\]

    • 2 years ago
  35. asnaseer Group Title
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    I /believe/ the original integral is not as straightforward as it might look. this is what wolfram thinks the answer should be: http://www.wolframalpha.com/input/?i=integrate+cos^2%28x%29*sqrt%28sin%28x%29%29

    • 2 years ago
  36. RedPrince Group Title
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    @asnaseer yes, you are right but how can I solve it i.e., \[\int\limits \sqrt{m-m^3}\]

    • 2 years ago
  37. asnaseer Group Title
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    see the link I posted - it is not an easy integral (even in this form): http://www.wolframalpha.com/input/?i=integrate+sqrt%28x-x^3%29

    • 2 years ago
  38. asnaseer Group Title
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    you need an expert like @JamesJ - he might be able to see a way to getting to the answer.

    • 2 years ago
  39. RedPrince Group Title
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    @asnaseer O.K., just see who help me in solving this question!

    • 2 years ago
  40. vamgadu Group Title
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    @RedPrince , the qustion is mis understood. the first solution is the correct one

    • 2 years ago
  41. asnaseer Group Title
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    @vamgadu have you read ALL the comments above? the original equation to solve was:\[∫\cos^2x \sqrt{sinx}dx\]this was incorrectly transformed to:\[\int\limits \sqrt{1-t^4}.2tdt\]

    • 2 years ago
  42. RedPrince Group Title
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    @FoolForMath help please!

    • 2 years ago
  43. RedPrince Group Title
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    @heena @amistre64 @AccessDenied @satellite73 help plaese if you have idea!

    • 2 years ago
  44. amistre64 Group Title
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    just integrate by parts

    • 2 years ago
  45. amistre64 Group Title
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    (1+t^4)^1/2 2t ..... hmm, seems i confused my integrating and deriving terms again -2

    • 2 years ago
  46. amistre64 Group Title
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    its still by parts but a trig sum does seem plausible

    • 2 years ago
  47. amistre64 Group Title
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    let tan(u) = t^2 sec^2(u) du = 2t dt replace the substitutions\[\int \sqrt{1+tan^2(u)}\ du\]

    • 2 years ago
  48. amistre64 Group Title
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    really need to wake up this morning :) \[\int \sqrt{1+tan^2(u)}\ sec^2(u)du\]

    • 2 years ago
  49. amistre64 Group Title
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    that would have worked out if I hadnt misread the operator; but same concept using the a sin or cos instead of a tangent

    • 2 years ago
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