Question

Help!!! Trigonometric Substitution! Integral (x/sqrt(x^2 - 4x + 8)

Answer 1

\[\int\limits \frac{ x }{ \sqrt{x^{2}-4x+8} }dx\]

Answer 2

I know that,
\[x=2\tan \theta + 2\]

Answer 3

\[dx=2\sec^2 \theta d \theta \]

Answer 4

I'm curious how you're approaching this.
I would start by completing the square in the denominator.
Makes it easier to see the substitution.
\[\Large \int\limits \frac{x}{\sqrt{(x-2)^2+4}}dx\]

Answer 5

Yeah! I did that, too!

Answer 6

ok cool :)

Answer 7

Hmm yah your sub looks good so far.
Just having trouble plugging everything in or what? :o

Answer 8

Yeah, that's my problem!

Answer 9

Let's get this last little piece of information:\[\Large x=2\tan \theta+2 \qquad\to\qquad (x-2)=2\tan \theta\]
I only placed the brackets there so it's easier to see where we're going to place it in the integral.

Answer 10

\[\large \int\limits\limits \frac{x}{\sqrt{(x-2)^2+4}}dx \qquad\to\qquad \int\limits\limits \frac{2\tan \theta+2}{\sqrt{(2\tan \theta)^2+4}}(2\sec^2\theta\;d \theta)\]

Answer 11

Understand how those are plugging in? :o
We had 3 pieces of information we had to use to make that substitution.

Answer 12

Yeah, so far, so good! :-)

Answer 13

\[\Large \int\limits \frac{2\tan \theta+2}{\sqrt{4\tan^2\theta+4}}(2\sec^2\theta \;d \theta)\]Hmm looks like we'll have some nice cancellations from here.

Answer 14

Okay, you cancel out \[4\tan^2 \theta \]

Answer 15

with \[2\sec^2 \theta\]

Answer 16

Right?

Answer 17

Sorry for the delay :( OpenStudy crashing again.. grr \[\large \int\limits\limits \frac{2\tan \theta+2}{\sqrt{4(\tan^2\theta+1)}}(2\sec^2\theta \;d \theta) \quad=\quad \int\limits\limits \frac{2\tan \theta+2}{\sqrt{4(\sec^2\theta)}}(2\sec^2\theta \;d \theta)\]\[\large =\int\limits \frac{2\tan \theta+2}{2\sec \theta}(2\sec^2\theta\;d \theta)\]

Answer 18

So it looks like it will only cancel out with one of the secants up top right?
Because we had to take the square root in the bottom, losing a power.

Answer 19

Yes! I think so, too!

Answer 20

Mmm ok that simplified down a bit for us!\[\Large \int\limits (2 \tan \theta+2)(\sec \theta\; d \theta)\]

Answer 21

Don't we have to do the right triangle?

Answer 22

At some point, yes we'll need to do that.
Whether now or later, it doesn't matter.

Answer 23

I think this is the relationship we'll want to draw, yes?\[\Large \tan \theta= \frac{x-2}{2}\]

Answer 24

Understand how to label this triangle? :)

Answer 25

I do! But for this problem in specific, I'm not sure how I will label it

Answer 26

\[\Large x=2\tan \theta+2 \qquad\to\qquad \tan \theta=\frac{x-2}{2}=\frac{opposite}{adjacent}\]

Answer 27

Go it! So the will the hypotenuse be \[\sqrt{x^2-4x+8}\]

Answer 28

?

Answer 29

ya looks good.

Answer 30

Okay, now, how do you get to the final answer?

Answer 31

We don't want to use the triangle just yet though! :O
Our integral is written in terms of trig functions (which will make the integration step much easier for us).

After we integrate, we can undo our substitutions by using the triangle.

Answer 32

\[\Large \int\limits\limits (2 \tan \theta+2)(\sec \theta\; d \theta) \quad=\quad 2\int\limits \sec \theta \tan \theta+ \sec \theta \;d \theta\]

Answer 33

Remember how to integrate those two terms or no? :D

Answer 34

I would be great if we can go over them again! I just want to make sure that I fully understand the concept!

Answer 35

It would*

Answer 36

There's not much to the first term, just something to remember.\[\Large (\sec \theta)' \quad=\quad \sec \theta \tan \theta\]
Does that look familiar? :o

Answer 37

Yes, it does!

Answer 38

Again for this other term, it's one that you'll simply want to memorize:\[\Large \int\limits \sec \theta\; d \theta \qquad=\qquad \ln\left|\sec \theta+\tan \theta\right|\]We can go through the steps if you want, but there's not a lot of intuition in solving it.

Answer 39

Okay, got it!

Answer 40

When do you we use the right triangle in order to arrive to the final answer?

Answer 41

The answer is supposed to be \[2\ln(\sqrt{x^2-4x+8}+x-2)+ \sqrt{x^2-4x+8}\] \]

Answer 42

So we integrated getting:\[\large 2\int\limits \sec \theta \tan \theta + \sec \theta\; d \theta \qquad=\qquad 2\left[\sec \theta+\ln\left(\sec \theta+\tan \theta\right)\right]+C\]

Answer 43

Now we can finally use our triangle.

Answer 44

We already have tangent, right? \(\Large \dfrac{x-2}{2}\)

We'll also need to figure out secant from the triangle we set up.

Answer 45

Still confused on that integration step? :3

Answer 46

No, I got it!

Answer 47

Proceed

Answer 48

based on our triangle we drew, it looks like:\[\large \sec \theta\quad=\quad \frac{\sqrt{x^2-4x+8}}{2}\]Hmm the square root bar is tweaking out on me -_- should be above the entire numerator there.

Answer 49

Okay!

Answer 50

So we just need to use those relationships to undo our substitutions now:\[\Large 2\color{#CC0033}{\sec \theta}+2\ln\left(\color{#CC0033}{\sec \theta}+\color{#3366CF}{\tan \theta}\right)+C\]

Answer 51

\[\Large 2\color{#CC0033}{\frac{\sqrt{x^2-4x+8}}{2}}+2\ln\left(\color{#CC0033}{\frac{\sqrt{x^2-4x+8}}{2}}+\color{#3366CF}{\tan \theta}\right)+C\]

Answer 52

Understand how to replace the tangent also?

Answer 53

I do! So, the 2's cancel out, and the final answer is \[2\ln (\sqrt{x^2-4x+8}+x-2)+\sqrt{x^2-4x+8}\]

Answer 54

Rigth?

Answer 55

yah the 2's cancel out in that term.
Where did the 2 inside the log go? :o hmm

Answer 56

We should have:\[\large 2\ln\left(\frac{\sqrt{x^2-4x+8}+x-2}{2}\right)+\sqrt{x^2-4x+8}+C\]

Answer 57

To match the answer in the book or whatever, we would need to apply another step.
We need to use a rule of logarithms.\[\large \log\left(\frac{a}{b}\right) \quad=\quad \log(a)-\log(b)\]

Answer 58

Oh, okay! But the answer we got is correct, right?

Answer 59

\[2\ln\left(\sqrt{x^2-4x+8}+x-2\right)\color{orangered}{-2\ln(2)}+\sqrt{x^2-4x+8}+\color{orangered}{C}\]
From here 2ln(2) is just a constant so we can absorb it into the C.

Answer 60

Yah our answer was correct :) No need to go any further unless you want to.

Answer 61

Makes it a little tricky when the books oversimplify these things lol

Answer 62

Oh wow! Really, THANK YOU SO MUCH for all the time you spend! I sincerely appreciate it! You were really clear and had patient! I now understand! THANK YOU! :-)

Answer 63

yay team \c:/ np!