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Substitution Rule

  1. Apr 22, 2007 #1
    Can someone please explain the Substitution Rule to me in easier terms? I am soooo confused!


  2. jcsd
  3. Apr 22, 2007 #2
    where are you consfused at? You are being vary vague, so us a sample problem and show us where you are confused at.
  4. Apr 22, 2007 #3
    Here are the problems that I have.

    View attachment 5.5.doc

    Thanks again,

  5. Apr 22, 2007 #4

    Gib Z

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    Well Ok Do you know the integrals of some general things?

    The Following are quite straightforward and require no substitution.

    [tex]\int \frac{x^2}{2} dx = ?[/tex]

    [tex]\int \sin x dx = ? [/tex]

    [tex]\int \frac{1}{x} dx =?[/tex]

    If You can do those, then its simple to explain how substitution helps. If you can't do those, its not as much a problem with substitution as it is with Fundamental Knowledge of Integration, such as the Fundamental Theorem Of Calculus.
  6. Apr 23, 2007 #5
    Yes, I know how to do those things. I mostly get confused when you have to write the problem with u an du.
  7. Apr 23, 2007 #6

    Gib Z

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    Well then there should be no problem, u is just another letter! We Could write u or y or a or l or anything. The first question, [tex]\int u du[/tex], you would be able to do [itex]\int x dx[/itex] so why not u du? If you want change the letter back to x for a second, but that could confuse you with the original x.

    [tex]\int x dx = x^2/2 + C[/tex]
    [tex]\int u du = u^2/2 +C[/tex]

    Or for the fourth one, [itex]\int \frac{dx}{x} = \ln |x| + C[/itex], so what is [itex]\int du/u[/itex]?
  8. Apr 23, 2007 #7
    I understand that u is just another letter, but after you take the hardest part of the equation and find the derivative you have some du=dx. Now you must write that in the form of the equation. This is the part where I get stuck.
  9. Apr 23, 2007 #8

    Gib Z

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    Ahh So sometimes you have trouble expressing the x's in terms of u?
  10. Apr 23, 2007 #9
    yeah substitution is really useful if you have a derivative inside the integral. Observe this integral: integral of (1/x)(1/lnx)dx , you see lnx, the derivative is inside it. what is the derivative of lnx? correct, it is (1/x). see..., so it is inside. You may now ask, why is this relavent?

    Ask yourself if this a straight forward integration? Can you do a simple antiderivative and be done with it? thats where substitution comes in, you use substitution to organize your thoughts so that when equations get too complex and you cannot do the 4 consequtive operations in your head without making a mistake, it is very useful.

    *Let U = lnx

    you make U equal an expession in the integral in that it has its derivative inside the integral. Lets make U = lnx . so in this case dU(which mean derivative of U) = (1/x)dx(!!always write down the dx, don't forget it!!! Important!!!) you have two identities and constants like one you see above or any other constants are constants, which mean their identies remain constant. so substitute for the identities and transform the equation in terms of U, Because (1/x)dx = dU and lnx = U, the equation becomes, integral of (1/U)dU

    integral of (1/U)dU <----------- look how easy it looks!!!!

    After you do the integral in terms of U, simply substitue the identity of U back to its original form, and Walla!! It was so easy wasn't it?

    That is all there is to it, you will find how valuable this knowledge is.
    Last edited: Apr 23, 2007
  11. Apr 23, 2007 #10

    Gib Z

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    Ok Now I looked through your document properly. The first time, I only looked at the bits where you said you got stuck. Now I see you have some other problems :P

    Now With substitution, we use it when we see 2 things in an integral. The first is a function. The second is its derivative. The aim is to simplify the expression to something even a bit easier, hopefully to something we recognize how to do. Even if its something subtle.

    EG [tex]\int \tan (x^5+5) x^4 dx[/tex]
    Now Look at this. The Function x^5 + 5 is in the integral, even though it isn't alone. And its even what is being "tan-ed". And its derivative isn't even in there! Or is it?...

    The derivative of x^5 + 5 is 5x^4. 5x^4 isn't in there, but x^4 is! So how can we make that into what we want? We multiply the Integrand ( The Function being integrated, in this case tan (x^5 +5) x^4 ) by 5. Now we can't just put in a constant factor in where we like and let it be. We have to balance the integral, by putting a factor of 1/5 outside the integral. Because as you should know, constant factors can be taken in and out of integrals.

    So with the new integral, [itex]1/5 \int \tan (x^5+5) 5x^4 dx[/itex], we know its equal to the original one because when we take the factor of 5 out, 5/5=1, same integral! Now its ready for substitution :)

    We see a function, in this case x^5+5, and its derivative in the integrand now. When we see such an occurence, let u = The Function, for this one x^5 +5 . That is because since its derivative is in there it simplies the whole integral is this fashion : [itex]\int \tan (u) \frac{du}{dx} dx[/itex].

    You can see why thats good, the dx's cancel out leaving you without any x's!

    Now its a simpler integral! [itex]\int \tan (u) du[/itex] Even if you don't know how to find the integral of tan, its simpler than the integral before.

    In fact theres an exercise for you, find the integral of tan. Remember tan x = (sin x)/(cos x). Rewrite tan in that way. Hint: What is the derivative of cos x?

    Now I Will take you through each question in your document step by step :)

    Question 1.
    [tex]\int \frac{(\ln x)^2}{x} dx[/tex]
    Well I see ln x in there, and its derivative 1/x.
    So let u = ln x, and therefore du/dx = 1/x, or du = (1/x) dx.
    So now putting those into the integral, it becomes [itex]\int u^2 \frac{du}{dx} dx[/itex]. The dx's cancel out, and the remaining integral is easy!

    [tex]\frac{u^3}{3} + C[/tex].

    Watch out though, we aren't done! The question wasn't in terms of u, we just made it so to make it easier to do! We have to replace the u with the x terms again. In this case u = ln x, so the integral is [tex]\frac{(\ln x)^3}{3} + C[/tex].

    Question 2.
    [tex]\int \sin^6 \theta \cos \theta d\theta[/tex]
    Now that one could fool some people. Cos is there, and its derivative -sin could be there if we balanced the factors of -1.

    However, since sin^6 is there, u=cos theta doesn't help Us. That is because we want its derivative, du/d(thtea), to cancel out with the d(theta)! But That won't happen since its sin^6, which is (du/dtheta)^6.

    So you do the other option, we see Sin, and its derivative cos is there all nice and alone.

    [itex]u = \sin \theta, du/d\theta=\cos \theta[/itex].

    Putting those into the integral, it becomes [tex]\int u^6 \frac{du}{d\theta} d\theta[/tex].
    dx's Cancel out, remanining integral is easy, u^7/7 +C, but remember to but back in terms of theta.
    [tex]\frac{(\sin \theta)^7}{7} +C[/tex]

    It didn't really matter if you made the wrong choice of substitution at the beginning, you can always start again when it doesn't work and try the other option. With time and experience you will usually get the best way first :)

    Question 3.
    [tex]\int \frac{z^2}{(1+z^3)^{\frac{1}{3}}} dz[/tex] (Cube root is the same as power 1/3).

    We see z^3 +1, and we almost see its derivative, 3z^2. So we put a factor of 3 inside the integral, balancing with a 1/3 outside.
    [tex]\frac{1}{3}\int \frac{1}{(1+z^3)^{\frac{1}{3}}}3z^2 dz[/tex].

    So now, u = z^3 +1, du/dz =3z^2.
    [tex]\frac{1}{3}\int \frac{1}{u^{1/3}} \frac{du}{dz} dz[/tex]

    dz's cancel out, and When you flip a fraction, its the same as to the power of -1. And when you have a power inside brackets, and then a power outside, you multiply the powers. So we get
    [tex]\frac{1}{3} \int u^{-1/3} du [/tex].

    To finish that off we use the "Power Rule for Integrals":
    [tex]\int x^n dx = \frac{x^{n+1}}{n+1} + C[/tex].

    Use the Power Rule for Integrals, [tex]\frac{1}{3} \int u^{-1/3} du = \frac{1}{3} \frac{u^{2/3}}{2/3} +C[/tex].

    Some simple algebra and not forgetting to replace u back into terms of z can simplify the integral to [tex](z^3+1)^{2/3}/2 +C[/tex].

    Question 4.
    [tex]\int \frac{1}{x} \frac{1}{\ln x} dx[/tex]

    We see ln x and its derivative!
    u=ln x, du/dx = 1/x
    [tex]\int \frac{1}{u} \frac{du}{dx} dx[/tex]
    dx's cancel, and you know the remaining Integral to be ln u +C!

    Hold on, replace u in terms of x again!
    [tex]\ln (\ln x) + C[/tex].
    Look at that, a log inside a log :)

    Now that we have finished those questions, here are some simple ones for you to do yourself:

    1. [tex]\int \sin^3 x \cos x dx[/tex]

    2. [tex] \int x (x^2+1)^{10} dx[/tex]

    3. [tex]\int 3x \sqrt{1+x^2} dx[/tex]

    4. [tex]\int \frac{x}{\sqrt{1-x^4}} dx[/tex]
    That ones a bit harder, so I tell you the substitution you use. u = x^2
    And use this rule, you will learn it later : [tex]\int \frac{1}{\sqrt{1-v^2}} dv = \arcsin v +C[/tex]. Remember v is just any letter.

    5. [tex]\int \frac{x+3}{\sqrt{x^2+6x-5}} dx[/tex]
    Hint: What do we see when we put in a factor of 2 in the integral?

    6. [tex]\int 4x e^{x^2} dx[/tex]

    7. [tex]\int \frac{\cos x}{(1+2\sin x)^2} dx[/tex]
    Hint: u = 1 + 2 sin x

    I Think thats more then enough, good luck!
    Last edited: Apr 23, 2007
  12. Apr 23, 2007 #11

    Gib Z

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    Ahh In the long long response I forgot about a bigger problem..You can't differentiate...the derivative of (ln x)^2 is not 2/x. If I don't post another tutorial by the time you read this, look up the chain rule in wikipedia or google and tell me what the derivative should be.

    And I just see now..Theres more pages in your document...Forget my exercises try the ones in your document but before that, learn the chain rule, product rule, and quotient rules for differentiation. I will post some exercises involving those tomorrow.
  13. Apr 23, 2007 #12
    I think an easier way to think about substitution is to "absorb" terms into the differential rather than do a magical substitution and change variables.

    [tex]\int (\cos x)^6 \sin x dx[/tex]
    Absorb sin x into the differential: sin x*dx = d(cos x). The general rule is f'(x)dx=df(x).
    [tex]\int (\cos x)^6 d(\cos x)[/tex]
    This is just a polynomial you're integrating! If you're getting lost, you could change variables here.
    [tex]\frac{(\cos x)^7}{7}[/tex]

    This makes it really easy to see a lot of substitutions, for example
    [tex]\int x e^{x^2} dx[/tex]
    Here, if you absorb x, you get x dx = 1/2*d(x^2), and x^2 is alone in the integral you're left with.
  14. Apr 23, 2007 #13
    I know the product, quotient, and chain rules. I get confused when I get to the something du= something dx. Now, I have to take that and put it in the original. Please explain that part. Is there like a formula or something for that part?
  15. Apr 23, 2007 #14


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    http://folk.ntnu.no/bronner/temp/temp1177371573.09375.png [Broken]
    Last edited by a moderator: May 2, 2017
  16. Apr 24, 2007 #15

    Gib Z

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    Well, du= something dx is the problem? Well after you know how to choose a proper u, finding du/dx should be no problem.

    If u = x^2, then du/dx = 2x.

    So once you have du/dx, just take the dx to the other side. Like I said, we want to see a function and its derivative. [itex]\int x^2 2x dx[/itex]. We see x^2 and 2x. So u = x^2, so we write the x^2 as u. And du/dx = 2x, du = 2x dx, so we replace the 2x dx with du, viola [itex]\int u du[/itex].
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