Here's the question you clicked on:
escolas
By completing the square and using a trig substitution, evaluate
\[\int\limits_{}^{}\frac{ x dx }{-x^{2}-2x+3 }\]
After completing the square I got \[\int\limits_{}^{}\frac{ x dx }{ (x+1)^2 +4 }\]
But I'm not sure what the mean by using a trig substitution...
Hmm did you complete the square correctly? I feel like there should be a negative on the outside..
\[\large -x^2-2x+3 \qquad = \qquad -(x^2+2x-3)\]Then to complete the square, we'll add and subtract 1.\[\large -(\color{orangered}{x^2+2x+1}-1-3) \qquad = \qquad -(\color{orangered}{(x+1)^2}-4)\]
See the negative confused me too, I forgot about that. that seems right
If you don't want to factor out the negative, then I guess we would have,\[\large 4-(x+1)^2\]
Which is probably the way we want to write it :)
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
But what does it mean by a trig substitution?
So are you familiar with the idea of `U Substitution`? It's a similar idea, just a little bit more complex c:
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?
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.
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.
Ok yea that makes sense I think..
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.\[\large a^2-x^2 \qquad = \qquad a^2-(a \sin \theta)^2 \qquad = \qquad a^2-a^2\sin^2 \theta\]\[\large a^2(1-\sin^2\theta) \qquad = \qquad a^2(\color{royalblue}{\cos^2\theta})\]
Lemme know if you're confused about any of that simplification. The blue step was where we applied this identity,\[\large \color{royalblue}{\cos^2\theta+\sin^2\theta=1}\qquad \rightarrow \qquad \color{royalblue}{\cos^2\theta=1-\sin^2\theta}\]
so then you have x over that though, where can you go from there?
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 \(\large x=a\sin\theta\).
oh nvm...you substitute the x in the numerator too...
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.
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.
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?
Let's write our integral like this, maybe it's easier to read this way,\[\large \int\limits\limits \frac{x}{4-(x+1)^2}\left(dx\right)\]
Oh i see what you're saying. :) Hmm we'll have to be a little sneaky I guess. \[\large x+1=2\sin \theta \qquad \rightarrow \qquad x=2\sin \theta -1\]
Again...a little slow. that makes perfect sense
Hold on the microwave beeped!! :O brb lol
So I think I correctly reduced it down to \[\int\limits_{}^{}\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }\]
Does that seem right so far?
\[\large \int\limits\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }(dx)\]Yah looks good so far :)
But I need a \[d \theta\]
haha is that just the derivative of x+1?
\[\large x+1=2\sin \theta\]We want to take the derivative of both sides, with respect to x. Yah good, both sides.
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.
\[\large dx=2\cos \theta\; d \theta\]That's what you came up with? yah looks good.
Ok cool this still isn't loooking too great haha
Oh do the cosines cancel out?
\[\large \int\limits\limits\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }(2\cos \theta \;d \theta)\]It looks like we can make a few cancellations.\[\large \int\limits\limits\limits\frac{ \cancel2\sin \theta-1}{ \cancel4\cos^{\cancel2\;1} \theta }(\cancel2\cancel{\cos \theta} \;d \theta)\]
the -1 is still messing with it though, isn't it?
\[\large \int\limits \frac{\sin \theta-1}{\cos \theta}d \theta\] Leaving us with this I believe.
Yes it is! But notice that the -1 is now on TOP!
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! :)
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?
Yes good, tan - sec i think.
But we're now in terms of theta rather than x, will it be hard to change back?
Which are actually... both... really terrible terms to integrate, you want to memorize both of these :) lol
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.
secx is secxtanx i think? but i can't remember tanx...
No, Integrating, secxtanx gives secx. Not the other way around :) We're going to get a couple of logs from both of these terms.
crap haha how do you integrate them then? are they just formulas?
Ah I found them in the back of my book
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.\[\large \int\limits \sec \theta d \theta \quad = \quad \ln|\sec \theta+\tan \theta|\]\[\large \int\limits \tan \theta d \theta \quad = \quad -\ln|\cos \theta|\]
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.
Are the absolute values gonna make things even more difficult?
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.
Ok cool, so now i solve for theta right?
\[\theta=\sin^{-1} \frac{ x+1 }{2 }\]
\[\int\limits \tan \theta - \sec \theta \; d \theta\] \[= \ln|\sec \theta|-\ln|\sec \theta+ \tan \theta|+C\] So this is what we came up with so far right?
Right! and does that make sense?
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.
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?
|dw:1359837750633:dw|\[x+1=2\sin \theta \qquad \rightarrow \qquad \sin \theta=\frac{x+1}{2} \qquad \rightarrow \qquad \sin \theta=\frac{opposite}{hypotenuse}\]
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..
I think we end with something likeeeeee\[\large \sqrt{2^2-(x+1)^2} \qquad = \qquad \sqrt{4-(x+1)^2}\]
Yep just wrote it out! So then I find the secant and tangent of that triangle?
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.
wow thats interesting....and i can just leave the answer like that, right?
well the logs can simplify to one over the other actually right?
Hmm yah you could combine the logs i suppose, but not much simplification is needed after you've put it back in x.
Just don't forget that +C! lol
Yea it'll probably make it more messy if anything
No prob \c:/ hopefully this all makes a bit more sense now!
it makes a ton more sense haha, sadly i never learned that so I was very confused when i saw those directions..