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Math Help - Substitution Rule for Integration

  1. #1
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    Substitution Rule for Integration

    I'm having trouble understanding what i'm doing here to get the answer.

    For instance

    Find indefinite integral for x^3cos((x^4)+2)dx

    ok first of all, what is the dx and why is it there at the end?

    To do this problem from the instructions, I let

    u = (x^4 ) + 2
    du = 4x^3 dx

    Is du just the derivative of u? Also why is dx there again and what does it represent?

    from here I would

    x^3 dx = (1/4)du

    Now that I have those, according to the substitution rule, it should be

    (cos u)((1/4)du) ok...now this is the part I don't understand even more is

    for Cos u, you find the antiderivative and get Sin u where u will be replaced with (x^4) + 2 giving you Sin((x^4) +2). That part is easily understandable. The part I don't understand is why is ((1/4)du) just = 1/4? where did the du go? U was replaced easily with ((x^4) +2) so why can't du be replaced with 4x^3 but instead it's replaced with 1/4? To my understanding, the Substitution Rule formula is f(u) du and this means f(u) times the derivative of u. so the equation should look like

    cos((x^4) + 2) multiplied by 4x^3. Apparently somewhere along in there i'm missing something and I don't get it. I'm interpreting the formula exactly the way it's written and can't get an understanding of what steps i'm skipping or missing.

    Any help is appreciated.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    I'm having trouble understanding what i'm doing here to get the answer.

    For instance

    Find indefinite integral for x^3cos((x^4)+2)dx

    ok first of all, what is the dx and why is it there at the end? <-- That means we are integrating with respect to x. Nothing special.

    To do this problem from the instructions, I let

    u = (x^4 ) + 2
    du = 4x^3 dx

    Is du just the derivative of u? Also why is dx there again and what does it represent?

    **********************
    u = (x^4 ) + 2 \rightarrow \frac{du}{dx} = 4x^3 \rightarrow du = 4x^3 dx
    **********************


    from here I would

    x^3 dx = (1/4)du

    Now that I have those, according to the substitution rule, it should be
    (cos u)((1/4)du) ok...now this is the part I don't understand even more is

    x^3 dx is now simply replaced by du AND the 1/4 should be taken to outside the integral. It is now a constant by which the final integral should be multiplied.
    x^4 is equal to u, and substituted accordingly

    for Cos u, you find the antiderivative and get Sin u where u will be replaced with (x^4) + 2 giving you Sin((x^4) +2). That part is easily understandable. The part I don't understand is why is ((1/4)du) just = 1/4? where did the du go? U was replaced easily with ((x^4) +2) so why can't du be replaced with 4x^3 but instead it's replaced with 1/4? To my understanding, the Substitution Rule formula is f(u) du and this means f(u) times the derivative of u. so the equation should look like

    cos((x^4) + 2) multiplied by 4x^3. Apparently somewhere along in there i'm missing something and I don't get it. I'm interpreting the formula exactly the way it's written and can't get an understanding of what steps i'm skipping or missing.

    Any help is appreciated.
    ..
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  3. #3
    Bar0n janvdl's Avatar
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    Also do not double post please. You also posted this here.

    See Rule #1 of the forum.
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  4. #4
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    sorry about the double post .

    x^3 dx is now simply replaced by du AND the 1/4 should be taken to outside the integral. It is now a constant by which the final integral should be multiplied.
    x^4 is equal to u, and substituted accordingly


    If x^3 dx is replaced by du then what happens to du when it gets to

    cos u times du?

    It just up and disappears and becomes

    (1/4) sin u + C

    I understand why cos u became sin u but i'm not understanding where did du disappear to.


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  5. #5
    Jen
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    du just tells you that you are taking the integral with respect to u.

    It is just instructions, it goes away after you take the integral
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  6. #6
    Bar0n janvdl's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    Find indefinite integral for x^3cos((x^4)+2)dx
     \int (x^3) cos(x^4 + 2) dx

    ************************** Start of Substitution ("u" calculations)

    Let \ u = x^4 + 2

    \frac{du}{dx} = 4x^3

    du = 4x^3 dx

    \frac{1}{4} du = x^3 dx

    ************************** (End "u" calcs)

    Let's rearrange a bit.

    \int cos(x^4 + 2) \ (x^3) dx <---- See the x^3 dx part? That's now \frac{1}{4} du. And we already said u = x^4 + 2

    \frac{1}{4} \int cos(u) \ du

    Which you can solve very easily. Just make sure to substitute x^4 + 2 = u back into the equation afterwards.
    Last edited by janvdl; April 18th 2008 at 02:33 PM. Reason: NOTE THE CHANGES I MADE... THE 1/4 AT THE END.
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  7. #7
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    Let's rearrange a bit.

    <---- See the part? That's now . And we already said



    Which you can solve very easily. Just make sure to substitute back into the equation afterwards.

    When it gets to that last line, that's where I get REALLY confused and can't understand why. Honestly, I can solve a good portion of my homework just following the examples but it's worthless to me unless I know why each thing is happening and for what purpose . So that last line, I interpret that as

    (Cos(u))(Derviative of u = 4x^3)

    So why isn't the problem written exactly as it's appearing which would be

    (Cos(u))(4x^3)

    ? That's what gets me
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  8. #8
    Bar0n janvdl's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    Let's rearrange a bit.

    <---- See the part? That's now . And we already said



    Which you can solve very easily. Just make sure to substitute back into the equation afterwards.

    When it gets to that last line, that's where I get REALLY confused and can't understand why. Honestly, I can solve a good portion of my homework just following the examples but it's worthless to me unless I know why each thing is happening and for what purpose . So that last line, I interpret that as

    (Cos(u))(Derviative of u = 4x^3)

    So why isn't the problem written exactly as it's appearing which would be

    (Cos(u))(4x^3)

    ? That's what gets me
    Just re-check my post. I forgot to add the  \frac{1}{4} parts at two places.
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  9. #9
    Jen
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    Quote Originally Posted by janvdl View Post

    Let's rearrange a bit.

    \int cos(x^4 + 2) \ (x^3) dx <---- See the x^3 dx part? That's now \frac{1}{4} du. And we already said u = x^4 + 2

    \frac{1}{4} \int cos(u) \ du

    Which you can solve very easily. Just make sure to substitute x^4 + 2 = u back into the equation afterwards.

    If you notice the final integral that Janvdl sets up has only u's in it. It has to end up this way because we can't end up with x's and u's, we wouldn't be able to integrate it. That is why we replace the x^3dx with the \frac{1}{4}du so that we can end up with only u's If you end up with both, you either did something wrong or a u du substitution wont work.

    You asked this question in the other thread
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  10. #10
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    MmMm...


    Ok how would I do this one

    4 / (1+2x)^3

    I would let u = 1 + 2x

    that would mean du = 2dx

    MMMmm...stuck
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  11. #11
    Jen
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    well 4=2*2

    so you can write this as

    \int \left (   \frac{2}{(1+2x)^3} \right )  2dx

    Then substitute what you found for u and du
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  12. #12
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    Quote Originally Posted by Jen View Post
    well 4=2*2

    so you can write this as

    \left ( \int  \frac{2}{(1+2x)^3}) \right  2dx

    Then substitute what you found for u and du

    so u can be 1 +2x

    and du can be 2(2dx)? = du =4dx?

    how would I do this problem from here? I haven't done a rational problem yet.
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  13. #13
    Jen
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    Quote Originally Posted by JonathanEyoon View Post
    so u can be 1 +2x

    and du can be 2(2dx)? = du =4dx?

    how would I do this problem from here? I haven't done a rational problem yet.
    when you found du it was 2dx right notice at the end of my integral (when i finally got the syntax right ) it shows a 2 dx

    So you can do direct substitution with du and leave the inside 2 there. (or pull it out in front of the integral, so you should get,

    2 \int \frac{1}{u^3}du
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  14. #14
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    ok, just to double check this is what I did


    u = 1 + 2x
    du = 2dx

    now here I got confused. From noticing a pattern and help form you guys, I noticed we always try to make the dx side the same as in the original integral. May I ask what's the reason for this?

    Well from where I left off

    (du= 2dx)(2)

    2du = 4dx

    from here

    2(1/u^3)

    (2u^(-3+1))/ (-3+1)

    -u^-2

    -1 / (1+2x)^2 + C

    Clap on or start over?
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  15. #15
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    ok, just to double check this is what I did


    u = 1 + 2x
    du = 2dx

    now here I got confused. From noticing a pattern and help form you guys, I noticed we always try to make the dx side the same as in the original integral. May I ask what's the reason for this?

    Well from where I left off

    (du= 2dx)(2)

    2du = 4dx

    from here

    2(1/u^3)

    (2u^(-3+1))/ (-3+1)

    -u^-2

    -1 / (1+2x)^2 + C

    Clap on or start over?
    I didnt think I saw anyone do it this way in the post...if there was I am sorry you have \int{x^3cos(x^4+2)dx}...by constant manipulation (i.e. multiplying by 1) we can change this to \frac{1}{4}\int{4x^3cos(x^4+2)dx}=\frac{sin(x^4+2)  }{4}+C...this is due to the fact that since by manipulation of the constants we got the integral into the form \int{f(g(x))\cdot{g'(x)}dx}...which basically means teh derivative of the inside of the function is outside and since the derivative of the quantity is 4x^3 if we get a four out front we can differentiate cos(x^4+2) as though it was cos(x) then replace the x with x^4+2...make sense?
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