A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

in Substitution Rule

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits\limits_{0}^{\frac{1}{2}}\frac{\sin^{-1}(x)}{\sqrt{1-x}} dx\]

  3. dumbcow
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i would make the substitution: x = sin^2(u) dx = 2sin*cos du \[\rightarrow 2\int\limits_{0}^{.5}u^{2} \sin(u) du\] Then use integration by parts

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but I don't know how to do it

  5. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Do you know integration by parts?

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah i do but that question is on substitution not in integration by parts

  7. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Why can't we do both?

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you can do :)

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it

  10. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits_{}^{}u^2 \sin(u) du=u^2(- \cos(u))-\int\limits_{}^{}2u (-\cos(u)) du\] you will need to do integration by parts one more time

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok Thank you a lot guys :)

  12. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    one thing I am confused on here, where did we get u^2 ? are you saying arcsin(sin^2u)=u^2 ?

  13. dumbcow
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes arcsin(sin u) = u by definition of the inverse function

  14. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    somehow the idea that that extends to arcsin(sin^n(u))=u^n has never been brought to my attention. thanks

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    could we have just used the substitution:\[u = \sin^{-1}(x)\]

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ah disregard, i didnt look at the denominator properly.

  17. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah i don't know if that is true turing

  18. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah it can't be

  19. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    arcsin(sin(u^n))=u^n arcsin(sin^n(u))=?

  20. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i agree with the first equation you have turing

  21. dumbcow
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    wait now that i look at it again, i think i have it backwards. sorry everyone sin^2(arcsin u) = u^2 , thats what i thought it was

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got

  23. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Oh man, I thought I was loosing it :/

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i got |dw:1327181038255:dw|

  25. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but i'm not sure about my answer

  26. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    wolfram disagrees http://www.wolframalpha.com/input/?i=integral+from+0+to+0.5+%28arcsin%28x%29%2Fsqrt%281-x%29 but I don't trust wolfram anymore, lol

  27. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't know man that question is hard to me :(

  28. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is it written correctly on the file you posted?

  29. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yeah, if there was an x^2 in the denom our lives would be about a million times easier

  30. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let me check it again, I may write it wrong

  31. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah you were right TuringTest I'm really sorry guys

  32. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i'm sorry guys there is X^2

  33. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}\]so \[u=\sin^{-1}x\to du=\frac{1}{\sqrt{1-x^2}}dx\]giving\[\int udu\]

  34. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah that what i did so far , but i wanna know if i'm doing right TuringTest. Can you do until the end of it ?

  35. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}}\]I'm going to be lazy and not use the proper bound notation, just plug numbers in at the end. \[u=\sin^{-1}x\to du=\frac{1}{\sqrt{1-x^2}}\]giving\[\int udu=\frac{u^2}{2}=\frac{1}{2}(\sin^{-1}x)^2|_{0}^{1/2}=\frac{1}{2}[(\sin^{-1}(1/2))^2-(\sin^{-1}(0))^2]\]\[=\frac{1}{2}[(\frac{\pi}{6})^2-0]=\frac{\pi^2}{72}\]

  36. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank You so much Turing that was very helpful

  37. TuringTest
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    anytime :)

  38. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.