Results 1 to 6 of 6

Math Help - Integral Substitution

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    3

    Integral Substitution

    Hi,
    Im trying to teach myself the substitution rule for integrals and am having a problem with the formula. I'm not grasping something here and hope someone can help me out.

    Here is the integral I'm working with.

    4x^3 * sqrt(x^4 + 7) dx

    The formula says f(g(x)g'(x)dx = Integral f(u) du

    To me that says to take f(u) * du, but when they work the problem out it seems that they just drop the du.

    If we start out with the integral

    4x^3 * sqrt(x^4 + 7) dx

    g(x) = u
    u = x^4 + 7
    du = 4x^3

    Now if I plug in the numbers following the formula I would write this out.

    (x^4 + 7)^1/2 * (4x^3)

    Then the examples goes on to say for the next step you end up with this

    2/3(u)^3/2 + c

    What happens to the 4x^3? From reading the formula I would think we have to multiple u and du? Any help is appreciated.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,115
    Thanks
    992
    Quote Originally Posted by jjstuart View Post
    Hi,
    Im trying to teach myself the substitution rule for integrals and am having a problem with the formula. I'm not grasping something here and hope someone can help me out.

    Here is the integral I'm working with.

    4x^3 * sqrt(x^4 + 7) dx

    The formula says f(g(x)g'(x)dx = Integral f(u) du

    To me that says to take f(u) * du, but when they work the problem out it seems that they just drop the du.

    If we start out with the integral

    4x^3 * sqrt(x^4 + 7) dx

    g(x) = u
    u = x^4 + 7
    du = 4x^3 dx
    note the correction above ...

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

    du = 4x^3 \, dx
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    You just need to write it out more and you'll get it! Here

    \displaystyle u = x^4 + 7
    \displaystyle du = 4x^3 dx

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

    Now, replace these in the integral to get

    \displaystyle \int 4x^3\sqrt{x^4 + 7} \;dx = \int 4x^3\sqrt{u} \frac{du}{4x^3} = \int \sqrt{u}\;du

    Since the 4x^3 factor cancels out!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jul 2010
    Posts
    3
    Thank you for the replies. I see how they cancel out now. I guess the hardest part for me too understand is how dx = du/4x3. I think I have more reading up to do on change of variables, I think?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    hello i think this should help you get started

    Before continuing learn table integrals (about 20-30 depending where someone is considering some of them to be or not to be table integrals)

    1 if you have function  f(x) so that

    \displaystyle \int f(x) \;dx = \int f(\varphi (t)) \varphi ' (t) \; dt

    where x= \varphi (t)

    it's basic way to solve it with substitution... (by which you will reduce your integral to some of the basic "table" integrals ) Use suitable substitution x=\varphi (t) of the new variable "t", you have to express everything with "t" and of course dx=\varphi ' (t) \;dt also if you don't succeed in that than you didn't chose good substitution (that is not a big problem... just go back and chose another one that all comes from much much practice )

    let's look few examples (simple ones... from those you start )


    1) example

    I=\int (x+1)^{100} \;dx

    substitute :

    (x+1) = t

    dx=dt

    I=\int (x+1)^{100} \;dx = \int t^{100} \; dt = \frac {t^{101}}{101} + C = \frac {(x+1)^{101}}{101} +C

    and much more of those but point is that integrals shaped like this :

    \int f(ax+b) \;dx

    using substitute (ax+b)=t you can simplify your integral so that :

    \int f(ax+b) \;dx =\frac {1}{a} \int f(t) \; dt


    ]integrals shaped like this :

    I= \int \frac {dx}{ax\pm b}

    using substitute

    (ax\pm b)=t

    dx=\frac{dt}{a}

    you can simplify your integral so that:

    I= \int \frac {dx}{ax\pm b}= \int \frac {\frac {dt}{a}}{t}=\frac {1}{a} \int \frac {dt}{t}=\frac {1}{a} ln |t| + C = \frac {1}{a} ln|ax\pm b| +C



    2) example

    I= \int \frac {dx}{1+4x^2} = \int \frac {dx}{1+(2x)^2}

    substitute

    2x=t \Rightarrow 2dx=dt \rightarrow dx=\frac{dt}{2}

    I= \int \frac {dx}{1+4x^2} = . . . . . = \frac {1}{2} \arctan {2x} + C



    3) example

    I= \int \frac {dx}{a^2+x^2 }

    this one is table one but...

    I= \int \frac {dx}{a^2+x^2 } = \int \frac {dx}{a^2(1+(\frac {x}{a})^2) } = \frac {1}{a^2} \int \frac {dx}{1+(\frac {x}{a})^2}

    substitute \frac {x}{a}=t \Rightarrow dx=a \; dt

    I= \int \frac {dx}{a^2+x^2 }= . . . .=\frac {1}{a} \arctan{\frac {x}{a}}  +C


    4) example

    I = \int \frac {dx}{\sqrt{x^2\pm a^2}}

    this one can be used as table integral ...

    substitute

    \sqrt{x^2\pm a^2}=t-x \Rightarrow \sqrt{x^2\pm a^2}+x=t

    ( \frac {x}{\sqrt{x^2\pm a^2} +1} )dx=dt \Rightarrow . . . \Rightarrow dx=\frac {t-x}{t} dt

    I = \int \frac {dx}{\sqrt{x^2\pm a^2}}= \int \frac {\frac {t-x}{t}}{t-x}\; dt = \int \frac {dt}{t} = \ln {t} +C = \ln {\sqrt{x^2\pm a^2}+x} +C


    anyway, there is much to be said about integration and methods of solving integrals. These up there are on substitution based Perhaps (sure) i miss few key examples (sorry about that, I will edit if it comes to my mind) But for get along with integrals you should practice (learn basic/table integrals, because you when solving integrals you are just finding the way to write your integral similar to any of those basic) substitution, partial integration... and to know basic types (because any integral is solved by the pattern )

    those are :

    \displaystyle I= \int \frac {P(x)}{Q(x)} \;dx

    \displaystyle I = \int R(x,\sqrt [n]{ax+b}) \;dx

    \displaystyle I= \int R(x,\sqrt [n]{\frac {ax+b}{cx+d}} )\;dx

    \displaystyle I= \int x^p (ax^q+b)^r \;dx

    \displaystyle I= \int \frac {dx}{ax^2+bx+c} \;dx

    \displaystyle I= \int \frac {dx}{\sqrt{ax^2+bx+c}} \;dx

    \displaystyle I= \int \sqrt{ax^2+bx+c} \;dx

    \displaystyle I= \int \frac {P_n(x)}{\sqrt{ax^2+bx+c}} \;dx

    \displaystyle I= \int P_n(x) \sqrt{ax^2+bx+c} \;dx

    \displaystyle I= \int R(\sin {x} , \cos {x} ) \; dx

    and so on... variation of those and perhaps few that I forgot to put them there but like i said, this to get you started good luck

    P.S. just practice and practice, that will make difference from good substitutions and bad ones
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jul 2010
    Posts
    3
    Thank you! It makes a lot more sense to me now. You're right, I just need to sit down and work out a lot more problems.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] U substitution with integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 14th 2011, 04:20 PM
  2. Integral by substitution
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 26th 2009, 01:22 PM
  3. def integral with substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 24th 2009, 07:39 PM
  4. Integral Substitution
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 19th 2008, 08:15 PM
  5. integral by substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 22nd 2008, 12:46 PM

Search Tags


/mathhelpforum @mathhelpforum