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dfresenius

  • 2 years ago

Trig integrals. sqrt(1-cos2x)dx

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  1. dfresenius
    • 2 years ago
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    |dw:1360952276270:dw|

  2. dfresenius
    • 2 years ago
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    should i do a u sub or change cos2x identity?

  3. JamesJ
    • 2 years ago
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    Well, cos(2x) = cos^2 x - sin^2 x = 1 - 2sin^2 x, hence \[ \sqrt{1 - \cos 2x} = \sqrt{1 - (1 - 2\sin^2 x)} \] which can be nicely simplified.

  4. dfresenius
    • 2 years ago
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    my answer is wrong

  5. JamesJ
    • 2 years ago
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    What does this expression simplify to? \[ \sqrt{1 - \cos 2x} = \sqrt{1 - (1 - 2\sin^2 x)} \]

  6. JamesJ
    • 2 years ago
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    \[ \sqrt{1 - (1 - 2\sin^2 x)} = \sqrt{2\sin^2 x} = ... \]

  7. dfresenius
    • 2 years ago
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    |dw:1360954273501:dw|

  8. JamesJ
    • 2 years ago
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    \[ \sqrt{1 - (1 - 2\sin^2 x)} = \sqrt{2\sin^2 x} = \sqrt{2} \sin x \] provided sin x > 0. Hence \[ \int \sqrt{1 - \cos 2x} \ dx \ = \ \int \sqrt{2} \sin x \ dx \ = \ ... \]

  9. dfresenius
    • 2 years ago
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    sqrt(2)-cosx

  10. dfresenius
    • 2 years ago
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    \[-\sqrt(2)cosx\]

  11. JamesJ
    • 2 years ago
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    + C, yes

  12. dfresenius
    • 2 years ago
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    from 0 to pi, the book asnwer says its 2sqrt2

  13. dfresenius
    • 2 years ago
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    i get 0

  14. JamesJ
    • 2 years ago
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    What is cos(pi) = ...? And cos(0) = ... ?

  15. dfresenius
    • 2 years ago
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    -1 and 1

  16. dfresenius
    • 2 years ago
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    sqrt(2) + sqrt(2) oh ya 2sqrt 2

  17. dfresenius
    • 2 years ago
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    sorry for wasting your time :(

  18. JamesJ
    • 2 years ago
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    np

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