Question

By completing the square and using a trig substitution, evaluate

Answer 1

\[\int\limits_{}^{}\frac{ x dx }{-x^{2}-2x+3 }\]

Answer 2

After completing the square I got \[\int\limits_{}^{}\frac{ x dx }{ (x+1)^2 +4 }\]

Answer 3

But I'm not sure what the mean by using a trig substitution...

Answer 4

Hmm did you complete the square correctly? I feel like there should be a negative on the outside..

Answer 5

\[\large -x^2-2x+3 \qquad = \qquad -(x^2+2x-3)\]Then to complete the square, we'll add and subtract 1.\[\large -(\color{orangered}{x^2+2x+1}-1-3) \qquad = \qquad -(\color{orangered}{(x+1)^2}-4)\]

Answer 6

See the negative confused me too, I forgot about that. that seems right

Answer 7

If you don't want to factor out the negative, then I guess we would have,\[\large 4-(x+1)^2\]

Answer 8

Which is probably the way we want to write it :)

Answer 9

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

Answer 10

But what does it mean by a trig substitution?

Answer 11

So are you familiar with the idea of `U Substitution`?
It's a similar idea, just a little bit more complex c:

Answer 12

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?

Answer 13

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.

Answer 14

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.

Answer 15

Yep, perfect!

Answer 16

Ok yea that makes sense I think..

Answer 17

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.\[\large a^2-x^2 \qquad = \qquad a^2-(a \sin \theta)^2 \qquad = \qquad a^2-a^2\sin^2 \theta\]\[\large a^2(1-\sin^2\theta) \qquad = \qquad a^2(\color{royalblue}{\cos^2\theta})\]

Answer 18

Lemme know if you're confused about any of that simplification.
The blue step was where we applied this identity,\[\large \color{royalblue}{\cos^2\theta+\sin^2\theta=1}\qquad \rightarrow \qquad \color{royalblue}{\cos^2\theta=1-\sin^2\theta}\]

Answer 19

wow thats awesome...

Answer 20

so then you have x over that though, where can you go from there?

Answer 21

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 \(\large x=a\sin\theta\).

Answer 22

oh nvm...you substitute the x in the numerator too...

Answer 23

I'm slow haha sorry

Answer 24

Yah good call :D

Answer 25

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.

Answer 26

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.

Answer 27

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?

Answer 28

Let's write our integral like this, maybe it's easier to read this way,\[\large \int\limits\limits \frac{x}{4-(x+1)^2}\left(dx\right)\]

Answer 29

Oh i see what you're saying. :)
Hmm we'll have to be a little sneaky I guess.
\[\large x+1=2\sin \theta \qquad \rightarrow \qquad x=2\sin \theta -1\]

Answer 30

Again...a little slow. that makes perfect sense

Answer 31

Hold on the microwave beeped!! :O brb lol

Answer 32

haha no worries!!

Answer 33

So I think I correctly reduced it down to \[\int\limits_{}^{}\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }\]

Answer 34

Does that seem right so far?

Answer 35

\[\large \int\limits\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }(dx)\]Yah looks good so far :)

Answer 36

But I need a \[d \theta\]

Answer 37

true story :O

Answer 38

haha is that just the derivative of x+1?

Answer 39

no 2sinx

Answer 40

\[\large x+1=2\sin \theta\]We want to take the derivative of both sides, with respect to x.
Yah good, both sides.

Answer 41

\[2\cos \theta\]?

Answer 42

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.

Answer 43

\[\large dx=2\cos \theta\; d \theta\]That's what you came up with? yah looks good.

Answer 44

Ok cool this still isn't loooking too great haha

Answer 45

XD

Answer 46

Oh do the cosines cancel out?

Answer 47

\[\large \int\limits\limits\frac{ 2\sin \theta-1}{ 4\cos ^2 \theta }(2\cos \theta \;d \theta)\]It looks like we can make a few cancellations.\[\large \int\limits\limits\limits\frac{ \cancel2\sin \theta-1}{ \cancel4\cos^{\cancel2\;1} \theta }(\cancel2\cancel{\cos \theta} \;d \theta)\]

Answer 48

the -1 is still messing with it though, isn't it?

Answer 49

\[\large \int\limits \frac{\sin \theta-1}{\cos \theta}d \theta\]
Leaving us with this I believe.

Answer 50

Yes it is! But notice that the -1 is now on TOP!

Answer 51

haha is that better?

Answer 52

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! :)

Answer 53

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?

Answer 54

Yes good, tan - sec i think.

Answer 55

But we're now in terms of theta rather than x, will it be hard to change back?

Answer 56

Which are actually... both... really terrible terms to integrate, you want to memorize both of these :) lol

Answer 57

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.

Answer 58

secx is secxtanx i think? but i can't remember tanx...

Answer 59

No, Integrating, secxtanx gives secx.

Not the other way around :)

We're going to get a couple of logs from both of these terms.

Answer 60

crap haha how do you integrate them then? are they just formulas?

Answer 61

Ah I found them in the back of my book

Answer 62

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.\[\large \int\limits \sec \theta d \theta \quad = \quad \ln|\sec \theta+\tan \theta|\]\[\large \int\limits \tan \theta d \theta \quad = \quad -\ln|\cos \theta|\]

Answer 63

Oh ok cool c:

Answer 64

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.

Answer 65

Are the absolute values gonna make things even more difficult?

Answer 66

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.

Answer 67

Ok cool, so now i solve for theta right?

Answer 68

\[\theta=\sin^{-1} \frac{ x+1 }{2 }\]

Answer 69

\[\int\limits \tan \theta - \sec \theta \; d \theta\]
\[= \ln|\sec \theta|-\ln|\sec \theta+ \tan \theta|+C\]
So this is what we came up with so far right?

Answer 70

Right! and does that make sense?

Answer 71

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.

Answer 72

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?

Answer 73

\[x+1=2\sin \theta \qquad \rightarrow \qquad \sin \theta=\frac{x+1}{2} \qquad \rightarrow \qquad \sin \theta=\frac{opposite}{hypotenuse}\]