Question

Integrate 1/sqrt(6x-x^2). (Stuck half way through)

Answer 1

I complete the square and get integral of \[\frac{ 1 }{ \sqrt{-(x-3)^{2}+9} }\]

Answer 2

the I do u-sub, u=x-3 \[\frac{ 1 }{ \sqrt{9-u ^{2}} }\]

Answer 3

I think I need to do trig sub where u=asin(theta) so u=3sin(theta)

Answer 4

but I'm not sure if i can do that because u is not x and I won't be able to substitute u back to x

Answer 5

Rewrite your \[\frac{ 1 }{ \sqrt{9-u ^{2}} }\] as \[\frac{ 1 }{ \sqrt{3^2-u ^{2}} }\] and note the similarility of this last result to \[\frac{ 1 }{ \sqrt{a^2-u ^{2}} }\]

Answer 6

Now you could either use trig substitution to integrate this, or find the appropriate integral in a table of integrals.

Answer 7

Just supposing that you had chosen u=3 cos theta, then \[\frac{ u }{ 3 }=\cos \theta \]

Answer 8

and so \[\theta=\cos ^{-1}\frac{ u }{ 3 }\]

Answer 9

no, I get that. What I'm not sure about is if integral of \[\int\limits_{?}^{?}\frac{ 1 }{\sqrt{9-(x-3)^{2}} } = \int\limits_{?}^{?} \frac{ 1 }{ \sqrt{9-(3\sin \theta}^2 }\]

Answer 10

* (3sin(theta))^2

Answer 11

sin theta should not yet be part of your integral. This substitution occurs AFTER integration. Please go thru the steps of integrating \[\int\limits\limits_{?}^{?}\frac{ 1dx }{\sqrt{9-(x-3)^{2}} } = \int\limits_{}^{}\frac{ 1 }{ \sqrt{3^2-u^2}du }\]

Answer 12

or simply look up this last integral in a table of integrals.

Answer 13

I guess I don't know/understand where to go from that last integral with the u-sub. What would be the next step?

Answer 14

Review the following:

\[\int\limits\limits\limits_{}^{}\frac{ 1dx }{\sqrt{9-(x-3)^{2}} } = \int\limits\limits_{}^{}\frac{ 1 }{ \sqrt{3^2-u^2} }du\]

Answer 15

Here we have temporarily dispensed with the function x-3, letting it be represented by u. Please look at the 2nd integral, above; are you to integrate this in terms of u, or to find the integral from a table?

Answer 16

Either or, it's a free response test review, as long as show all my work

Answer 17

\[\int\limits\limits_{}^{}\frac{ 1 }{ \sqrt{3^2-u^2} }du\] has the form\[ \int\limits\limits_{}^{}\frac{ 1 }{ \sqrt{a^2-u^2}}du\]

Answer 18

Please find a table of integrals now and determine the result of this last integral. You will either need to memorize this integral or know how to evaluate it via trig substitution.

Answer 19

ohhhhhh, so it would equal arcsin(x/a)

Answer 20

Need to label your result. What would equal arcsin (x/a)?

Answer 21

I was missing that table of integrals part, *smacks head*

Answer 22

so, your arcsin (x/a) represents what?

Answer 23

Don't overthink this; if you have a table of integrals in front of you, the table should give you the appropriate label for arcsin (x/a).

Answer 24

hint: arcsin (x/a) represents an ANGLE.

Answer 25

so arcsin(u/3) right?

Answer 26

You need to label "arcsin (x/a)." The correct label is the character 'theta.'

\[\theta = \arcsin (x/a)\]

Answer 27

Important: keep in mind that "arcsin y" is an angle: arcsin y is the angle whose sine is (y/1).

Answer 28

Let's step back for a moment to review where we are? Is this discussion shedding light on the material at hand or onlyl adding to your confusion?

Answer 29

What do you need to know to finish the solution of this problem?

Answer 30

I got everything up to how we got arcsin, now I need to find the value that represents arcsin right? So I know arcsin(u/3) but I don't quiet know how to find its value

Answer 31

OK. If you begin with arcsin (u/3), and you remember you have defined u as x-3, can you now rewrite arcsin (u/3) in terms of x?

Answer 32

ok, so arcsin((x-3)/3)

Answer 33

Looks good. You now have an integral, and that integral is in terms of x.

Answer 34

Again, please be sure to label each and every result.

Answer 35

oh wait, I though we had taken care of the integral already and integrated to get arcsin

Answer 36

Your task was to Integrate 1/sqrt(6x-x^2).

You could label Integrate 1/sqrt(6x-x^2) as "dy" and then go thru all the steps discussed here to find the result "y."

Thus, you could write "y=arcsin ([x-3] / 3) + c

Answer 37

Yes. A quick look at my table of integrals confirms that.

Answer 38

You could check your work thru differentiating your answer and determining whether it could be proved equal to 1/sqrt(6x-x^2).

Answer 39

ohhh, ok, I finally got it

Answer 40

Note how you begin with the variable x and you end with the variable x. If your intermediate result has theta or u in it, you are not finished.

Answer 41

The key here is that you must be able to recognize the appropriate trig substitution for each case. If you have Sqrt (a^2 - x^2) in the denom. of your integrand, then the appropriate substitution is what?

Answer 42

Again, refer to your table of integrals.

Answer 43

it would be arcsin(x/a), or is it x=asin(theat)

Answer 44

arcsin is the integrated answer but sin(theat) is the substitution, righ

Answer 45

As you can see, this discussion has been heavy on inverse trig functions and the algebraic concept of substitution. The correct subst. would be x=a sin theta.

Answer 46

In your current problem, a=3.

Answer 47

So the approp. subst. would be x=3 sin theta.

Answer 48

Note that it is fair to expect of you that you can use appropriate trig substitutions (in this case, inverse trig functions) to evaluate integrals such as that with which you originally started. Thus, you'd have at least two good tools in your toolbox: 1) table of integrals and 2) actual trig substitutions.

Answer 49

but in the problem we just worked out, we didn't substitute to sin(theta), we just used the table of integration to find arcsin

Answer 50

oh ok

Answer 51

so I can use either one

Answer 52

That's right. We / you used the table of integrals.

I'd suggest you look up "trig substitutions" in your textbook and review this material.


For the problem at hand,

Answer 53

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

Answer 54

you need to be able to recognize that the correct trig subst. is x=3 sin theta.

Answer 55

if x = 3 sin theta, then

x^2 = 9 (sin theta)^2

You must also find dx in terms of theta and d(theta).

Look familiar or not so familiar?

Answer 56

yes, I recognize this, which is why I was so lost the way we worked it out, I hadn't looked at the table

Answer 57

Altho I need to excuse myself now, I'd be happy to continue working with you if you find you still need some guidance. Tag me later on if you wish to continue, but remember that there are others on OpenStudy who could also help you competently.

Answer 58

ok, thank you so much!

Answer 59

My habit is to reach for my Calculus textbook whenever I'm not 100% sure I understand and/or remember the tools I need to solve a given problem.

You're welcome. Best of luck.