anonymous
  • anonymous
Integration help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1442609775803:dw|
anonymous
  • anonymous
@Empty
anonymous
  • anonymous
i have the answer as 9.747 rounded to 3 decimal places but i dont understand how to get

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Zarkon
  • Zarkon
\[x=\frac{\tan(\theta)}{2}\]
anonymous
  • anonymous
????
anonymous
  • anonymous
i need explanation please
zepdrix
  • zepdrix
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..
zepdrix
  • zepdrix
Our problem involves something of the form: \(\rm 1+stuff^2\) Which is telling us that we want to use the tangent identity, ya?
zepdrix
  • zepdrix
\[\large\rm \int\limits\sqrt{1+4x^2}~dx=\int\limits\sqrt{1+(2x)^2}~dx\]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!
zepdrix
  • zepdrix
So, as Zarkon indicated, make the substitution, \(\large\rm 2x=\tan\theta\)
zepdrix
  • zepdrix
So we're making the substitution: \(\large\rm \color{orangered}{2x=\tan\theta}\) Then our integral becomes:\[\large\rm \int\limits\limits\sqrt{1+(\color{orangered}{2x})^2}~dx=\int\limits\limits\sqrt{1+(\color{orangered}{\tan\theta})^2}~dx\]
zepdrix
  • zepdrix
We need to also replace the dx with something involving d(theta) though.
zepdrix
  • zepdrix
Whatchu think mage? :O Too confusing?

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