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Homework Help: Substitution Rule in Indefinite Integrals

  1. Jul 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Please explain how to use the substitution rule in indefinite integrals. I am unable even to start the problem.

    2. Relevant equations



    3. The attempt at a solution
     

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  2. jcsd
  3. Jul 22, 2010 #2
    Actually, one second. I am solving it right now. Will post a solution.
     
  4. Jul 22, 2010 #3

    Mark44

    Staff: Mentor

    Substitution is the reverse of the chain rule in differentiation. For example,
    [tex]\frac{d}{dx}(x^3 - 2)^4~=~4(x^3 - 2)^3~3x^2[/tex]

    The corresponding indefinite integral is
    [tex]\int 12x^2(x^3 - 2)^3~dx[/tex]
    Here, we let u = x3 - 2, so du = 3x2~dx, so
    [tex]\int 12x^2(x^3 - 2)^3~dx~=~\int 4*3x^2~(x^3 - 2)^3~dx~=~\int 4 u^3 du~=~\frac{4u^4}{4} + C~=~(x^3 - 2)^4 + C[/tex]

    Can you see how this technique (it's not a rule) could be applied to your problem?
     
  5. Jul 22, 2010 #4
    Thank you for the helpful note. I did not know that substitution is reverse of chain rule. Here is my solution. Correct?
     

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  6. Jul 22, 2010 #5

    Mark44

    Staff: Mentor

    Correct as far as it goes. You still need to undo your substitution.

    Also, u = x2 + 1 ==> du = 2xdx. You have x and dx, so you can get the 2 that's needed by multiplying by 2 and then multiplying on the outside of the integral by 1/2. That way, you're really multiplying by 1, which doesn't change the value of the integrand.
     
  7. Jul 22, 2010 #6
    Why do you put du=2xdx? I think we need dx=du/2x.

    Here is the completed version.
     

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  8. Jul 22, 2010 #7
    Can you check another problem please too? Thank you.
     

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  9. Jul 22, 2010 #8

    Mark44

    Staff: Mentor

    Sure, that works.

    Another substitution that isn't so obvious is to let u = sec2t (using t for theta). Then du = 2sec(2t)tan(2t)dt

    [tex]\int sec(2t)tan(2t)dt = \frac{1}{2}\int 2sec(2t)tan(2t)dt = \frac{1}{2}\int du= \frac{u}{2} + C = \frac{sec(2t)}{2} + C[/tex]
     
  10. Jul 22, 2010 #9
    Thank you very much!

    I am also stuck on this one. ex has been always tricky for me.
     

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  11. Jul 22, 2010 #10

    OH! This one is much prettier than mine!
     
  12. Jul 22, 2010 #11

    Mark44

    Staff: Mentor

    This will work, but will require another substitution.

    Here's a different approach: u = ex + 1, du = exdx
    [tex]\int \frac{e^x~dx}{e^x + 1}=\int \frac{du}{u}= ln(u) + C = ln(e^x+ 1) + C[/tex]

    Since u = ex + 1 > 0 for all real x, I didn't need to include absolute values on the argument to the natural log function.
     
  13. Jul 22, 2010 #12
    Thank you a lot, Mark44!
     
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