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TransendentialPI

  • 3 years ago

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

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  1. TransendentialPI
    • 3 years ago
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  2. cwrw238
    • 3 years ago
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    try u = sqrt(2x)

  3. TransendentialPI
    • 3 years ago
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    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
    • 3 years ago
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    I did find this one, but it is making the integral more complicated:

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  5. TransendentialPI
    • 3 years ago
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    That is not getting me to one of those MC answers/

  6. waterineyes
    • 3 years ago
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    Now use Integration By parts..

  7. waterineyes
    • 3 years ago
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    \[\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1 - x^2}}\]

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

  9. waterineyes
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Welcome dear..

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