Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

TransendentialPI

  • 2 years ago

I can't see the substitution I'm supposed to make. Any hints? See next for the exercise.

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

    1 Attachment
  2. cwrw238
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    try u = sqrt(2x)

  3. TransendentialPI
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I did |dw:1343307585147:dw||dw:1343307655512:dw| I'm supposed to be geting this into the form so I can find the formula in a integral table.

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

    I did find this one, but it is making the integral more complicated:

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

    That is not getting me to one of those MC answers/

  6. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Now use Integration By parts..

  7. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1 - x^2}}\]

  8. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int\limits u \cdot \sin^{-1}(u) du = \sin^{-1}(u) \cdot \frac{u^2}{2} - \int\limits( \frac{1}{\sqrt{1 - u^2}} \times \frac{u^2}{2}) du\] Try to solve for last part..

  9. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you can do it like this: \[ \frac{1}{\sqrt{1 - u^2}} \times u^2 = \frac{1 - u^2 + 1}{\sqrt{1 - u^2}} =\sqrt{1 - u^2} + \frac{1}{\sqrt{1 - u^2}}\] Use the following two Special Integral Formula: \[\int\limits \sqrt{a^2 - x^2}dx = \frac{1}{2}x \cdot \sqrt{a^2 - x^2} + \frac{1}{2}a^2 \cdot \sin^{-1} (\frac{x}{a}) + C\] \[\int\limits \frac{1}{\sqrt{a^2 - u^2}}dx = \sin^{-1}(\frac{x}{a}) + C\]

  10. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Sorry - will be there in between terms: \[\frac{1}{\sqrt{1 - u^2}} \times u^2 = \frac{1 - u^2 - 1}{\sqrt{1 - u^2}} =\sqrt{1 - u^2} - \frac{1}{\sqrt{1 - u^2}}\] By this your integration middle term will become positive: \[\int\limits\limits u \cdot \sin^{-1}(u) du = \sin^{-1}(u) \cdot \frac{u^2}{2} + \int\limits\limits( \frac{1}{\sqrt{1 - u^2}} \times \frac{-u^2}{2}) du\]

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

    Thanks, I just flipped back in the textbook section and they provide an example in the form \[\int\limits xsin^{-1} x \] dx= that integral and then to use a the table again. Thank so much!

  12. waterineyes
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Welcome dear..

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

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.