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anonymous
 3 years ago
By completing the square and using a trig substitution, evaluate
anonymous
 3 years ago
By completing the square and using a trig substitution, evaluate

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But I'm not sure what the mean by using a trig substitution...

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

zepdrix
 3 years ago
Best 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.\[\large (\color{orangered}{x^2+2x+1}13) \qquad = \qquad (\color{orangered}{(x+1)^2}4)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0See the negative confused me too, I forgot about that. that seems right

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Which is probably the way we want to write it :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What 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
 3 years ago
Best ResponseYou've already chosen the best response.0But what does it mean by a trig substitution?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.2Here 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
 3 years ago
Best ResponseYou've already chosen the best response.2The 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ok yea that makes sense I think..

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2It 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.\[\large a^2x^2 \qquad = \qquad a^2(a \sin \theta)^2 \qquad = \qquad a^2a^2\sin^2 \theta\]\[\large a^2(1\sin^2\theta) \qquad = \qquad a^2(\color{royalblue}{\cos^2\theta})\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Lemme 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
 3 years ago
Best ResponseYou've already chosen the best response.0so then you have x over that though, where can you go from there?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2So 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh nvm...you substitute the x in the numerator too...

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2In 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
 3 years ago
Best ResponseYou've already chosen the best response.2We 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ohh 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
 3 years ago
Best ResponseYou've already chosen the best response.2Let'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
 3 years ago
Best ResponseYou've already chosen the best response.2Oh 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
 3 years ago
Best ResponseYou've already chosen the best response.0Again...a little slow. that makes perfect sense

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Hold on the microwave beeped!! :O brb lol

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does that seem right so far?

zepdrix
 3 years ago
Best 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 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But I need a \[d \theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha is that just the derivative of x+1?

zepdrix
 3 years ago
Best 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.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Err 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
 3 years ago
Best 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok cool this still isn't loooking too great haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh do the cosines cancel out?

zepdrix
 3 years ago
Best 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.\[\large \int\limits\limits\limits\frac{ \cancel2\sin \theta1}{ \cancel4\cos^{\cancel2\;1} \theta }(\cancel2\cancel{\cos \theta} \;d \theta)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 1 is still messing with it though, isn't it?

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Yes it is! But notice that the 1 is now on TOP!

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2When 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
 3 years ago
Best ResponseYou've already chosen the best response.0that 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
 3 years ago
Best ResponseYou've already chosen the best response.2Yes good, tan  sec i think.

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

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Yes 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
 3 years ago
Best ResponseYou've already chosen the best response.0secx is secxtanx i think? but i can't remember tanx...

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0crap haha how do you integrate them then? are they just formulas?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah I found them in the back of my book

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2I 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
 3 years ago
Best ResponseYou've already chosen the best response.2the 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
 3 years ago
Best ResponseYou've already chosen the best response.0Are the absolute values gonna make things even more difficult?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2No, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Ok cool, so now i solve for theta right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\theta=\sin^{1} \frac{ x+1 }{2 }\]

zepdrix
 3 years ago
Best 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right! and does that make sense?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Let'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
 3 years ago
Best ResponseYou've already chosen the best response.0Ah 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
 3 years ago
Best ResponseYou've already chosen the best response.2dw: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
 3 years ago
Best ResponseYou've already chosen the best response.0then 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
 3 years ago
Best ResponseYou've already chosen the best response.2I think we end with something likeeeeee\[\large \sqrt{2^2(x+1)^2} \qquad = \qquad \sqrt{4(x+1)^2}\]

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Notice 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wow thats interesting....and i can just leave the answer like that, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well the logs can simplify to one over the other actually right?

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

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Just don't forget that +C! lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yea it'll probably make it more messy if anything

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah thank you so much!

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2No prob \c:/ hopefully this all makes a bit more sense now!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it 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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