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escolasBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}\frac{ x dx }{x^{2}2x+3 }\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
After completing the square I got \[\int\limits_{}^{}\frac{ x dx }{ (x+1)^2 +4 }\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
But I'm not sure what the mean by using a trig substitution...
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Hmm did you complete the square correctly? I feel like there should be a negative on the outside..
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\large x^22x+3 \qquad = \qquad (x^2+2x3)\]Then to complete the square, we'll add and subtract 1.\[\large (\color{orangered}{x^2+2x+1}13) \qquad = \qquad (\color{orangered}{(x+1)^2}4)\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
See the negative confused me too, I forgot about that. that seems right
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
If you don't want to factor out the negative, then I guess we would have,\[\large 4(x+1)^2\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Which is probably the way we want to write it :)
 one year ago

escolasBest ResponseYou've already chosen the best response.0
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
 one year ago

escolasBest ResponseYou've already chosen the best response.0
But what does it mean by a trig substitution?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
So are you familiar with the idea of `U Substitution`? It's a similar idea, just a little bit more complex c:
 one year ago

escolasBest ResponseYou've already chosen the best response.0
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?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Ok yea that makes sense I think..
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
It makes more sense when you have sqrt's in the bottom. But this problem still applies. When we see this form, \[\large a^2x^2\]We'll make the substitution \(\large x=a\sin\theta\). Here is what will happen when we do that.\[\large a^2x^2 \qquad = \qquad a^2(a \sin \theta)^2 \qquad = \qquad a^2a^2\sin^2 \theta\]\[\large a^2(1\sin^2\theta) \qquad = \qquad a^2(\color{royalblue}{\cos^2\theta})\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Lemme know if you're confused about any of that simplification. The blue step was where we applied this identity,\[\large \color{royalblue}{\cos^2\theta+\sin^2\theta=1}\qquad \rightarrow \qquad \color{royalblue}{\cos^2\theta=1\sin^2\theta}\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
so then you have x over that though, where can you go from there?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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 \(\large x=a\sin\theta\).
 one year ago

escolasBest ResponseYou've already chosen the best response.0
oh nvm...you substitute the x in the numerator too...
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
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?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Let's write our integral like this, maybe it's easier to read this way,\[\large \int\limits\limits \frac{x}{4(x+1)^2}\left(dx\right)\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Oh i see what you're saying. :) Hmm we'll have to be a little sneaky I guess. \[\large x+1=2\sin \theta \qquad \rightarrow \qquad x=2\sin \theta 1\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Again...a little slow. that makes perfect sense
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Hold on the microwave beeped!! :O brb lol
 one year ago

escolasBest ResponseYou've already chosen the best response.0
So I think I correctly reduced it down to \[\int\limits_{}^{}\frac{ 2\sin \theta1}{ 4\cos ^2 \theta }\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Does that seem right so far?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\large \int\limits\frac{ 2\sin \theta1}{ 4\cos ^2 \theta }(dx)\]Yah looks good so far :)
 one year ago

escolasBest ResponseYou've already chosen the best response.0
But I need a \[d \theta\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
haha is that just the derivative of x+1?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\large x+1=2\sin \theta\]We want to take the derivative of both sides, with respect to x. Yah good, both sides.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\large dx=2\cos \theta\; d \theta\]That's what you came up with? yah looks good.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Ok cool this still isn't loooking too great haha
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Oh do the cosines cancel out?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\large \int\limits\limits\frac{ 2\sin \theta1}{ 4\cos ^2 \theta }(2\cos \theta \;d \theta)\]It looks like we can make a few cancellations.\[\large \int\limits\limits\limits\frac{ \cancel2\sin \theta1}{ \cancel4\cos^{\cancel2\;1} \theta }(\cancel2\cancel{\cos \theta} \;d \theta)\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
the 1 is still messing with it though, isn't it?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\large \int\limits \frac{\sin \theta1}{\cos \theta}d \theta\] Leaving us with this I believe.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Yes it is! But notice that the 1 is now on TOP!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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! :)
 one year ago

escolasBest ResponseYou've already chosen the best response.0
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?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Yes good, tan  sec i think.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
But we're now in terms of theta rather than x, will it be hard to change back?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Which are actually... both... really terrible terms to integrate, you want to memorize both of these :) lol
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
secx is secxtanx i think? but i can't remember tanx...
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
No, Integrating, secxtanx gives secx. Not the other way around :) We're going to get a couple of logs from both of these terms.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
crap haha how do you integrate them then? are they just formulas?
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Ah I found them in the back of my book
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.\[\large \int\limits \sec \theta d \theta \quad = \quad \ln\sec \theta+\tan \theta\]\[\large \int\limits \tan \theta d \theta \quad = \quad \ln\cos \theta\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Are the absolute values gonna make things even more difficult?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Ok cool, so now i solve for theta right?
 one year ago

escolasBest ResponseYou've already chosen the best response.0
\[\theta=\sin^{1} \frac{ x+1 }{2 }\]
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
\[\int\limits \tan \theta  \sec \theta \; d \theta\] \[= \ln\sec \theta\ln\sec \theta+ \tan \theta+C\] So this is what we came up with so far right?
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Right! and does that make sense?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
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?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
dw:1359837750633:dw\[x+1=2\sin \theta \qquad \rightarrow \qquad \sin \theta=\frac{x+1}{2} \qquad \rightarrow \qquad \sin \theta=\frac{opposite}{hypotenuse}\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
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..
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
I think we end with something likeeeeee\[\large \sqrt{2^2(x+1)^2} \qquad = \qquad \sqrt{4(x+1)^2}\]
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Yep just wrote it out! So then I find the secant and tangent of that triangle?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
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.
 one year ago

escolasBest ResponseYou've already chosen the best response.0
wow thats interesting....and i can just leave the answer like that, right?
 one year ago

escolasBest ResponseYou've already chosen the best response.0
well the logs can simplify to one over the other actually right?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Hmm yah you could combine the logs i suppose, but not much simplification is needed after you've put it back in x.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
Just don't forget that +C! lol
 one year ago

escolasBest ResponseYou've already chosen the best response.0
Yea it'll probably make it more messy if anything
 one year ago

zepdrixBest ResponseYou've already chosen the best response.2
No prob \c:/ hopefully this all makes a bit more sense now!
 one year ago

escolasBest ResponseYou've already chosen the best response.0
it makes a ton more sense haha, sadly i never learned that so I was very confused when i saw those directions..
 one year ago
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