unimatix
  • unimatix
Integral 1/(x^2 sqrt(4-x^2)) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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SolomonZelman
  • SolomonZelman
\(x=2\sin\theta\)
SolomonZelman
  • SolomonZelman
@unimatix do you want a regular example where this method (trigonometric substitution) is used?
unimatix
  • unimatix
That would be great honestly.

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SolomonZelman
  • SolomonZelman
Ok, I am assuming you are familiar with a regular u-substitution, so I won't go to deep into explaining it.... \(\large\color{black}{ \displaystyle \int \frac{4}{\sqrt{1+x^2}}dx}\) \(\large\color{gray}{ \displaystyle x=\tan\theta}\) The derivative of \(\tan\theta\) is \(\sec^2\theta\), so your \(dx\) component will be: \(\large\color{gray}{ \displaystyle dx=\sec^2\theta{~} d\theta}\) \(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{1+\left(\tan\theta\right)^2}}d\theta}\) \(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{1+\tan^2\theta}}d\theta}\) \(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sqrt{\sec^2\theta}}d\theta}\) \(\large\color{black}{ \displaystyle \int \frac{4\sec^2\theta}{\sec\theta}d\theta}\) \(\large\color{black}{ \displaystyle \int 4\sec\theta{~}d\theta}\) and then apply the formula for integral of secant
SolomonZelman
  • SolomonZelman
you will then substitute the x back in this example, but what I wanted to show is this mechanism of substituting a trignonometric function instead of x into the integral.
SolomonZelman
  • SolomonZelman
Do you want to come back to our original problem?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{x^2\sqrt{4-x^2}}~dx}\) Try to do this substitution: \(\large \color{gray}{x=2\sin\theta}\) TELL ME: what is the derivative of \(2\sin\theta\) ?
SolomonZelman
  • SolomonZelman
(derivative with respect to \(\theta\) )
unimatix
  • unimatix
I can follow what you did after you substituted in. But I can't understand what made you decide to do the trig substitution there.
SolomonZelman
  • SolomonZelman
What made me decide is that I want to make substitution in a way that the part in ssquare root containing x^2+C would become just a single trigonometric function inside the square root.
unimatix
  • unimatix
Derivative should be 2cos right?
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
2cos\(\theta\)
unimatix
  • unimatix
Yeah that's what I meant, sorry
unimatix
  • unimatix
so I set u = 2sin(theta) and du = 2cos(theta) dx right?
SolomonZelman
  • SolomonZelman
There is a rule for trigonometric substitution ... it will get intuitive after some time, but for now - the rule: ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ FORM: (\(a^2\) is just a constant) SUBSTITUTION \(\Large x^2+a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\tan\theta \) \(\Large x^2-a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sec\theta \) \(\Large a^2-x^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sin\theta \) ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\color{blue}{x}^2\sqrt{4-\color{blue}{x}^2}}~\color{royalblue }{dx}}\) Try to do this substitution: \(\large \color{blue}{x=2\sin\theta}\) \(\large \color{red }{dx=2\cos\theta ~d \theta }\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\left(\color{blue}{2\sin\theta}\right)^2\sqrt{4-\left(\color{blue}{2\sin\theta}\right)^2}}~\color{red }{2\cos\theta ~d \theta}}\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}\)
SolomonZelman
  • SolomonZelman
When you are done reading tell me whether you have followed everything so far, or there are so parts you don't get.
unimatix
  • unimatix
Okay just worked through all the steps you went through
SolomonZelman
  • SolomonZelman
Ok, and now, you can tell this: \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4\left( 1-\sin^2\theta \right)}} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\cos\theta}{4\sin^2\theta \sqrt{\cos^2\theta }} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\cos\theta}{4\sin^2\theta \cos\theta } ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}\)
SolomonZelman
  • SolomonZelman
are you getting it?
unimatix
  • unimatix
Yeah I get that.
SolomonZelman
  • SolomonZelman
Can you finish the problem up?
SolomonZelman
  • SolomonZelman
What function has a derivative of csc²(\(\theta\)) ?
unimatix
  • unimatix
Is it cot(theta)?
unimatix
  • unimatix
or I think -cot(theta)
SolomonZelman
  • SolomonZelman
YES, -cot(\(\theta\))
unimatix
  • unimatix
-1/4 * cot(theta) ?
unimatix
  • unimatix
- 1/(4 tan(theta)) ?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}\) \(\large\color{black}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\csc^2\theta ~d \theta}\) \(\large\color{black}{\displaystyle-\frac{1}{4} \cot(\theta)+{\rm C} }\) and then you had that \(\large x=2\sin\theta \) so put back for x. \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\cos\theta}{\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\cos\theta}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\cos^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle\frac{-\sqrt{4-x^2}}{4x}+{\rm C} }\)
SolomonZelman
  • SolomonZelman
it was kind of long to put in for x.... And the check is: http://www.wolframalpha.com/input/?i=derivative+of+%28-%E2%88%9A%284-x%C2%B2%29%29%2F%284x%29
unimatix
  • unimatix
Sorry it's taking me a little while to go through it.
SolomonZelman
  • SolomonZelman
I just rearranged the answer in terms of 2sin(\(\theta\)), and substituted for x.
unimatix
  • unimatix
I'm confused as to how you got the third to last step
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C}=-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\)
SolomonZelman
  • SolomonZelman
Are you referring to that?
unimatix
  • unimatix
Yeah
unimatix
  • unimatix
I think I've figured it out. That's tricky though.
unimatix
  • unimatix
The last line is the final step right?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C}=}\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{blue}{\displaystyle-\frac{1}{2} \frac{\sqrt{\left(1/4 \right){~}(4){~}\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{(1/4)}\cdot \sqrt{{~}4{~}\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{(1/2)\cdot \sqrt{{~}4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{{~}4-\left(2\sin\theta\right)^2}}{2\sin\theta}+{\rm C} }\)
SolomonZelman
  • SolomonZelman
By the way, I made a mistake: \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }\) \(\large\color{red}{\displaystyle\frac{-\sqrt{4-x^2}}{8x}+{\rm C} }\)
SolomonZelman
  • SolomonZelman
the red is correct.
unimatix
  • unimatix
Oh I should've seen that.
unimatix
  • unimatix
|dw:1441678847661:dw|
unimatix
  • unimatix
ahh the dx should be 2 cos(theta)
SolomonZelman
  • SolomonZelman
well, that is how I dx should be: `2cos(theta) d(theta)`
SolomonZelman
  • SolomonZelman
and that is exactly how I started to substitute \(2\sin\theta\) for \(x\).
unimatix
  • unimatix
Oh sorry. I don't know what I was thinking . We just finished it.
unimatix
  • unimatix
Thank you so much. I'm going to go try and work through it a second time on my own. THANK YOU. THANK YOU. THANK YOU.
SolomonZelman
  • SolomonZelman
Sure, remember the formas though ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ FORM: (\(a^2\) is just a constant) SUBSTITUTION \(\Large x^2+a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\tan\theta\) \(\Large x^2-a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sec\theta\) \(\Large a^2-x^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sin\theta\) ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼ Like 4+x², is 2²+x² ---> form 1, x=atan(\(\theta\)).
unimatix
  • unimatix
Will do!
SolomonZelman
  • SolomonZelman
Ok, yw!

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