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anonymous

  • one year ago

Integration help

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  1. anonymous
    • one year ago
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    |dw:1442609775803:dw|

  2. anonymous
    • one year ago
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    @Empty

  3. anonymous
    • one year ago
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    i have the answer as 9.747 rounded to 3 decimal places but i dont understand how to get

  4. Zarkon
    • one year ago
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    \[x=\frac{\tan(\theta)}{2}\]

  5. anonymous
    • one year ago
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    ????

  6. anonymous
    • one year ago
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    i need explanation please

  7. zepdrix
    • one year ago
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    It's hard to take the root of a sum... that addition sign is causing a problem. We would like to apply some type of trigonometric substitution because our trig identities allow us to get rid of addition/subtraction. \(\rm 1-\sin^2x=\cos^2x\) \(\rm 1+\tan^2x=\sec^2x\) And the others..

  8. zepdrix
    • one year ago
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    Our problem involves something of the form: \(\rm 1+stuff^2\) Which is telling us that we want to use the tangent identity, ya?

  9. zepdrix
    • one year ago
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    \[\large\rm \int\limits\sqrt{1+4x^2}~dx=\int\limits\sqrt{1+(2x)^2}~dx\]We would like to turn the 2x into tangent, then we would have 1+tangent^2 under the root, and would be able to apply our identity!

  10. zepdrix
    • one year ago
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    So, as Zarkon indicated, make the substitution, \(\large\rm 2x=\tan\theta\)

  11. zepdrix
    • one year ago
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    So we're making the substitution: \(\large\rm \color{orangered}{2x=\tan\theta}\) Then our integral becomes:\[\large\rm \int\limits\limits\sqrt{1+(\color{orangered}{2x})^2}~dx=\int\limits\limits\sqrt{1+(\color{orangered}{\tan\theta})^2}~dx\]

  12. zepdrix
    • one year ago
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    We need to also replace the dx with something involving d(theta) though.

  13. zepdrix
    • one year ago
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    Whatchu think mage? :O Too confusing?

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