Integral 1/(x^2 sqrt(4-x^2)) dx

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Integral 1/(x^2 sqrt(4-x^2)) dx

Mathematics
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\(x=2\sin\theta\)
@unimatix do you want a regular example where this method (trigonometric substitution) is used?
That would be great honestly.

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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
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.
Do you want to come back to our original problem?
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{x^2\sqrt{4-x^2}}~dx}\) Try to do this substitution: \(\large \color{gray}{x=2\sin\theta}\) TELL ME: what is the derivative of \(2\sin\theta\) ?
(derivative with respect to \(\theta\) )
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.
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.
Derivative should be 2cos right?
yes
2cos\(\theta\)
Yeah that's what I meant, sorry
so I set u = 2sin(theta) and du = 2cos(theta) dx right?
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 \(\Large x^2+a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\tan\theta \) \(\Large x^2-a^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sec\theta \) \(\Large a^2-x^2{~~~~~~~~~~~~}\Rightarrow {~~~~~~~~~~~~}x=a\sin\theta \) ☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼☼
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\color{blue}{x}^2\sqrt{4-\color{blue}{x}^2}}~\color{royalblue }{dx}}\) Try to do this substitution: \(\large \color{blue}{x=2\sin\theta}\) \(\large \color{red }{dx=2\cos\theta ~d \theta }\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\left(\color{blue}{2\sin\theta}\right)^2\sqrt{4-\left(\color{blue}{2\sin\theta}\right)^2}}~\color{red }{2\cos\theta ~d \theta}}\)
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}\)
When you are done reading tell me whether you have followed everything so far, or there are so parts you don't get.
Okay just worked through all the steps you went through
Ok, and now, you can tell this: \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4-4\sin^2\theta}} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\sqrt{4\left( 1-\sin^2\theta \right)}} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}\)
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{2\cos\theta}{4\sin^2\theta\cdot (2)\cdot \sqrt{\cos^2\theta }} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\cos\theta}{4\sin^2\theta \sqrt{\cos^2\theta }} ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{\cos\theta}{4\sin^2\theta \cos\theta } ~d \theta}\) \(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}\)
are you getting it?
Yeah I get that.
Can you finish the problem up?
What function has a derivative of csc²(\(\theta\)) ?
Is it cot(theta)?
or I think -cot(theta)
YES, -cot(\(\theta\))
-1/4 * cot(theta) ?
- 1/(4 tan(theta)) ?
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{4\sin^2\theta } ~d \theta}\) \(\large\color{black}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\csc^2\theta ~d \theta}\) \(\large\color{black}{\displaystyle-\frac{1}{4} \cot(\theta)+{\rm C} }\) and then you had that \(\large x=2\sin\theta \) so put back for x. \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\cos\theta}{\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\cos\theta}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\cos^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle\frac{-\sqrt{4-x^2}}{4x}+{\rm C} }\)
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
Sorry it's taking me a little while to go through it.
I just rearranged the answer in terms of 2sin(\(\theta\)), and substituted for x.
I'm confused as to how you got the third to last step
\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C}=-\frac{1}{4} \frac{\sqrt{4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\)
Are you referring to that?
Yeah
I think I've figured it out. That's tricky though.
The last line is the final step right?
\(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{1-\sin^2\theta}}{2\sin\theta}+{\rm C}=}\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{blue}{\displaystyle-\frac{1}{2} \frac{\sqrt{\left(1/4 \right){~}(4){~}\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{\sqrt{(1/4)}\cdot \sqrt{{~}4{~}\left(1-\sin^2\theta \right)}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{2} \frac{(1/2)\cdot \sqrt{{~}4-4\sin^2\theta}}{2\sin\theta}+{\rm C} }\) \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{{~}4-\left(2\sin\theta\right)^2}}{2\sin\theta}+{\rm C} }\)
By the way, I made a mistake: \(\large\color{black}{\displaystyle-\frac{1}{4} \frac{\sqrt{4-(2\sin\theta)^2}}{2\sin\theta}+{\rm C} }\) \(\large\color{red}{\displaystyle\frac{-\sqrt{4-x^2}}{8x}+{\rm C} }\)
the red is correct.
Oh I should've seen that.
|dw:1441678847661:dw|
ahh the dx should be 2 cos(theta)
well, that is how I dx should be: `2cos(theta) d(theta)`
and that is exactly how I started to substitute \(2\sin\theta\) for \(x\).
Oh sorry. I don't know what I was thinking . We just finished it.
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.
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(\(\theta\)).
Will do!
Ok, yw!

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