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eistein+newton+gallileo=_me

  • 3 years ago

find the integral of 1/root(-x^2 + 6x - 8) with explanations

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  1. LagrangeSon678
    • 3 years ago
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    thats the integral?

  2. Izzy25
    • 3 years ago
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    i got the information from here: :/ idk tho http://mathforum.org/library/drmath/view/64571.html

  3. nikvist
    • 3 years ago
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    \[\int\frac{1}{\sqrt{-x^2+6x-8}}dx=\int\frac{1}{\sqrt{-x^2+6x-9+1}}dx=\int\frac{1}{\sqrt{1-(x-3)^2}}dx=\] \[=\arcsin{(x-3)}+C\]

  4. eistein+newton+gallileo=_me
    • 3 years ago
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    no the answer is the arcsin one by nikvist

  5. dumbcow
    • 3 years ago
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    Use completing the square \[\rightarrow \int\limits_{}^{} \frac{dx}{\sqrt{1-(x-3)^{2}}}\] ...yeah thats what i got

  6. eistein+newton+gallileo=_me
    • 3 years ago
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    CAN YOU PLEASE EXPLAIN

  7. Izzy25
    • 3 years ago
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    he wants you to explain

  8. nikvist
    • 3 years ago
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    \[\int\frac{1}{\sqrt{1-x^2}}dx=\arcsin{x}+C\] basic integral

  9. dumbcow
    • 3 years ago
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    u = x-3 du = dx \[\rightarrow \int\limits_{}^{} \frac{du}{\sqrt{1-u^{2}}}\] u = sin(theta) du = cos(theta) \[\rightarrow \int\limits_{}^{}\frac{\cos(\theta)}{\sqrt{1-\sin^{2}(\theta)}}d \theta = \int\limits_{}^{}\frac{\cos(\theta)}{\sqrt{\cos^{2}(\theta)}}d \theta = \int\limits_{}^{}d \theta = \theta+C\] \[\theta = \sin^{-1}(u) = \sin^{-1}(x-3)\]

  10. eistein+newton+gallileo=_me
    • 3 years ago
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    so automatically once i see such a question i should use trig substitutions ? what about this one 4/(4-x^2)

  11. LagrangeSon678
    • 3 years ago
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    integral for arc sin is \[\int\limits_{}^{} du/\sqrt{a^2-u^2}\]

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