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eistein+newton+gallileo=_me

  • 3 years ago

find the integral of 1/root(x^2-4)

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  1. myininaya
    • 3 years ago
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    |dw:1315245910188:dw|

  2. myininaya
    • 3 years ago
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    let \[\sec(\theta)=\frac{x}{2}\]

  3. myininaya
    • 3 years ago
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    \[2 \sec(\theta)=x => 2\sec(\theta)\tan(\theta) d \theta=dx\]

  4. myininaya
    • 3 years ago
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    \[\int\limits_{}^{}\frac{1}{\sqrt{(2\sec(\theta))^2-4}} 2\sec(\theta) \tan(\theta) d \theta \]

  5. myininaya
    • 3 years ago
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    \[\frac{2}{\sqrt{4}}\int\limits_{}^{}\frac{\sec(\theta) \tan(\theta)}{\sqrt{\sec^2(\theta)-1}} d \theta \]

  6. eistein+newton+gallileo=_me
    • 3 years ago
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    mmmmm the answer is ln[x + root(X^2 -4) ] + c

  7. eistein+newton+gallileo=_me
    • 3 years ago
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    mmmmm the answer is ln[x + root(X^2 -4) ] + c .......negative 4

  8. myininaya
    • 3 years ago
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    \[\int\limits_{}^{}\frac{\sec(\theta) \tan(\theta)}{\sqrt{\tan^2(\theta)}} d \theta\]

  9. myininaya
    • 3 years ago
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    \[\int\limits_{}^{}\sec(\theta) d \theta =\int\limits_{}^{}\sec(\theta) \cdot \frac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)} d \theta\] let \[u=\sec(\theta)+\tan(\theta)=>du=[\sec(\theta)\tan(\theta)+\sec^2(\theta)] d \theta\] so we have \[\int\limits_{}^{}\frac{du}{u}=\ln|u|+C\]

  10. myininaya
    • 3 years ago
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    but \[u=\sec(\theta)+\tan(\theta)\]

  11. myininaya
    • 3 years ago
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    so we have \[\ln|\sec(\theta)+\tan(\theta)|+C\] but we made the substition that \[\sec(\theta)=\frac{x}{2}\]

  12. myininaya
    • 3 years ago
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    so if \[\sec(\theta)=\frac{x}{2}\] |dw:1315246444736:dw| so \[\tan(\theta)=\frac{\sqrt{x^2-4}}{2}\]

  13. myininaya
    • 3 years ago
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    so we actually have \[\ln|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}|+C\]

  14. myininaya
    • 3 years ago
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    \[\ln|\frac{x+\sqrt{x^2-4}}{2}|+C=\ln|x+\sqrt{x^2-4}|-\ln|2|+C\]

  15. myininaya
    • 3 years ago
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    but ln(2) is just a constant so it can all be included in the C part so we could write \[\ln|x+\sqrt{x^2-4}|+K\]

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