anonymous
  • anonymous
By completing the square and using a trig substitution, evaluate
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\int\limits_{}^{}\frac{ x dx }{-x^{2}-2x+3 }\]
anonymous
  • anonymous
After completing the square I got \[\int\limits_{}^{}\frac{ x dx }{ (x+1)^2 +4 }\]
anonymous
  • anonymous
But I'm not sure what the mean by using a trig substitution...

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zepdrix
  • zepdrix
Hmm did you complete the square correctly? I feel like there should be a negative on the outside..
zepdrix
  • zepdrix
\[\large -x^2-2x+3 \qquad = \qquad -(x^2+2x-3)\]Then to complete the square, we'll add and subtract 1.\[\large -(\color{orangered}{x^2+2x+1}-1-3) \qquad = \qquad -(\color{orangered}{(x+1)^2}-4)\]
anonymous
  • anonymous
See the negative confused me too, I forgot about that. that seems right
zepdrix
  • zepdrix
If you don't want to factor out the negative, then I guess we would have,\[\large 4-(x+1)^2\]
zepdrix
  • zepdrix
Which is probably the way we want to write it :)
anonymous
  • anonymous
What I had originally done was moved the 3 over to the other side and then divided by -1 but I don't think thats right
anonymous
  • anonymous
But what does it mean by a trig substitution?
zepdrix
  • zepdrix
So are you familiar with the idea of `U Substitution`? It's a similar idea, just a little bit more complex c:
anonymous
  • anonymous
I am, is it something like the proof of the area of a circle where you randomly (or seemingly randomly) substitute in a sinx or cosx to magically make things work?
zepdrix
  • zepdrix
Here is our integral so far, \[\large \int\limits \frac{x\;dx}{4-(x+1)^2}\] Just to make sure we're on the same page.
zepdrix
  • zepdrix
The purpose of a `Trig sub` is that by applying one, we can get rid of the `addition/subtraction` in the bottom. We can turn it into a single term in the denominator, and it makes it a ton easier to deal with.
anonymous
  • anonymous
Yep, perfect!
anonymous
  • anonymous
Ok yea that makes sense I think..
zepdrix
  • zepdrix
It makes more sense when you have sqrt's in the bottom. But this problem still applies. When we see this form, \[\large a^2-x^2\]We'll make the substitution \(\large x=a\sin\theta\). Here is what will happen when we do that.\[\large a^2-x^2 \qquad = \qquad a^2-(a \sin \theta)^2 \qquad = \qquad a^2-a^2\sin^2 \theta\]\[\large a^2(1-\sin^2\theta) \qquad = \qquad a^2(\color{royalblue}{\cos^2\theta})\]
zepdrix
  • zepdrix
Lemme know if you're confused about any of that simplification. The blue step was where we applied this identity,\[\large \color{royalblue}{\cos^2\theta+\sin^2\theta=1}\qquad \rightarrow \qquad \color{royalblue}{\cos^2\theta=1-\sin^2\theta}\]
anonymous
  • anonymous
wow thats awesome...
anonymous
  • anonymous
so then you have x over that though, where can you go from there?
zepdrix
  • zepdrix
So we'll have to replace several pieces :) We'll also have to substitute in for the `dx`, so we'll have to take the derivative of our \(\large x=a\sin\theta\).
anonymous
  • anonymous
oh nvm...you substitute the x in the numerator too...
anonymous
  • anonymous
I'm slow haha sorry
zepdrix
  • zepdrix
Yah good call :D
zepdrix
  • zepdrix
In the problem we're working on, it's just a tiny bit more complicated, \(x\) is not the thing being squared. \((x+1)\) is. So THAT is the thing we'll setting up to substitute.
zepdrix
  • zepdrix
We want to let,\[\large x+1=2\sin \theta\]Take a look at that a minute, let it sink in :D Our \(a\) is 2 right? Because we have \(2^2\) in front of the subtraction.
anonymous
  • anonymous
Ohh ok yea I was just about to do it and got a little stuck there. that makes more sense. So how do we deal with the numerator then?
zepdrix
  • zepdrix
Let's write our integral like this, maybe it's easier to read this way,\[\large \int\limits\limits \frac{x}{4-(x+1)^2}\left(dx\right)\]
zepdrix
  • zepdrix
Oh i see what you're saying. :) Hmm we'll have to be a little sneaky I guess. \[\large x+1=2\sin \theta \qquad \rightarrow \qquad x=2\sin \theta -1\]
anonymous
  • anonymous
Again...a little slow. that makes perfect sense
zepdrix
  • zepdrix
Hold on the microwave beeped!! :O brb lol
anonymous
  • anonymous
haha no worries!!
anonymous
  • anonymous
So I think I correctly reduced it down to \[\int\limits_{}^{}\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }\]
anonymous
  • anonymous
Does that seem right so far?
zepdrix
  • zepdrix
\[\large \int\limits\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }(dx)\]Yah looks good so far :)
anonymous
  • anonymous
But I need a \[d \theta\]
zepdrix
  • zepdrix
true story :O
anonymous
  • anonymous
haha is that just the derivative of x+1?
anonymous
  • anonymous
no 2sinx
zepdrix
  • zepdrix
\[\large x+1=2\sin \theta\]We want to take the derivative of both sides, with respect to x. Yah good, both sides.
anonymous
  • anonymous
\[2\cos \theta\]?
zepdrix
  • zepdrix
Err maybe we want to take it with respect to theta, then we don't have the chain rule. It won't matter either way, I'm just trying to think about what will be easier to understand.
zepdrix
  • zepdrix
\[\large dx=2\cos \theta\; d \theta\]That's what you came up with? yah looks good.
anonymous
  • anonymous
Ok cool this still isn't loooking too great haha
zepdrix
  • zepdrix
XD
anonymous
  • anonymous
Oh do the cosines cancel out?
zepdrix
  • zepdrix
\[\large \int\limits\limits\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }(2\cos \theta \;d \theta)\]It looks like we can make a few cancellations.\[\large \int\limits\limits\limits\frac{ \cancel2\sin \theta-1}{ \cancel4\cos^{\cancel2\;1} \theta }(\cancel2\cancel{\cos \theta} \;d \theta)\]
anonymous
  • anonymous
the -1 is still messing with it though, isn't it?
zepdrix
  • zepdrix
\[\large \int\limits \frac{\sin \theta-1}{\cos \theta}d \theta\] Leaving us with this I believe.
zepdrix
  • zepdrix
Yes it is! But notice that the -1 is now on TOP!
anonymous
  • anonymous
haha is that better?
zepdrix
  • zepdrix
When it's in the bottom, that causes a real problem. When it's in the TOP, we can just split it up into a couple of fractions! :)
anonymous
  • anonymous
that makes sense...so you'd end up with the integral of tanx and the integral of secx? Well not exactly but pretty much, right?
zepdrix
  • zepdrix
Yes good, tan - sec i think.
anonymous
  • anonymous
But we're now in terms of theta rather than x, will it be hard to change back?
zepdrix
  • zepdrix
Which are actually... both... really terrible terms to integrate, you want to memorize both of these :) lol
zepdrix
  • zepdrix
Yes it will be a little bit tricky to change back, it will require us to setup a triangle and do some quick side work. It's not too bad though.
anonymous
  • anonymous
secx is secxtanx i think? but i can't remember tanx...
zepdrix
  • zepdrix
No, Integrating, secxtanx gives secx. Not the other way around :) We're going to get a couple of logs from both of these terms.
anonymous
  • anonymous
crap haha how do you integrate them then? are they just formulas?
anonymous
  • anonymous
Ah I found them in the back of my book
zepdrix
  • zepdrix
I don't really wanna go through the steps of integrating them right now XD It'll take too much time lol. So yes, for now let's just refer to the formulas.\[\large \int\limits \sec \theta d \theta \quad = \quad \ln|\sec \theta+\tan \theta|\]\[\large \int\limits \tan \theta d \theta \quad = \quad -\ln|\cos \theta|\]
zepdrix
  • zepdrix
Oh ok cool c:
zepdrix
  • zepdrix
the tangent can also be written as \[\large \ln|\sec \theta|\]By bringing the negative in as a power. Depending on which way you want to write it.
anonymous
  • anonymous
Are the absolute values gonna make things even more difficult?
zepdrix
  • zepdrix
No, I don't think we need to worry about that. We'll include the bars in our final answer but that's the jist of it. If we had a definite integral, and were plugging values in, we might want to be extra careful, but not in this case.
anonymous
  • anonymous
Ok cool, so now i solve for theta right?
anonymous
  • anonymous
\[\theta=\sin^{-1} \frac{ x+1 }{2 }\]
zepdrix
  • zepdrix
\[\int\limits \tan \theta - \sec \theta \; d \theta\] \[= \ln|\sec \theta|-\ln|\sec \theta+ \tan \theta|+C\] So this is what we came up with so far right?
anonymous
  • anonymous
Right! and does that make sense?
zepdrix
  • zepdrix
Let's not do that, if we had a lonesome \(\theta\) somewhere in our solution, then we would have to include the arcsine function. But since we have only trig functions in our solution, we'll do some triangle math.
anonymous
  • anonymous
Ah ok, I just figured we could get the angles theta was with triangles and then take the tan and sec of those! What do we need to do?
zepdrix
  • zepdrix
|dw:1359837750633:dw|\[x+1=2\sin \theta \qquad \rightarrow \qquad \sin \theta=\frac{x+1}{2} \qquad \rightarrow \qquad \sin \theta=\frac{opposite}{hypotenuse}\]
anonymous
  • anonymous
then use pythagorean to find that that side is well i'm too dumb to do that in my head actually. haha i have to write it down..
zepdrix
  • zepdrix
I think we end with something likeeeeee\[\large \sqrt{2^2-(x+1)^2} \qquad = \qquad \sqrt{4-(x+1)^2}\]
anonymous
  • anonymous
Yep just wrote it out! So then I find the secant and tangent of that triangle?
zepdrix
  • zepdrix
Notice how that value under the square root actually matches the denominator we started with? That will always happen.. If I'm remembering correctly. You should always see your initial substitution.. thing, in the triangle somewhere.
zepdrix
  • zepdrix
Yes good :)
anonymous
  • anonymous
wow thats interesting....and i can just leave the answer like that, right?
anonymous
  • anonymous
well the logs can simplify to one over the other actually right?
zepdrix
  • zepdrix
|dw:1359838191896:dw|
zepdrix
  • zepdrix
Hmm yah you could combine the logs i suppose, but not much simplification is needed after you've put it back in x.
zepdrix
  • zepdrix
Just don't forget that +C! lol
anonymous
  • anonymous
Yea it'll probably make it more messy if anything
anonymous
  • anonymous
ah thank you so much!
zepdrix
  • zepdrix
No prob \c:/ hopefully this all makes a bit more sense now!
anonymous
  • anonymous
it makes a ton more sense haha, sadly i never learned that so I was very confused when i saw those directions..

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