Substitution Rule in Indefinite Integrals

  • Thread starter phillyolly
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Homework Statement



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

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The Attempt at a Solution

 

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  • #2
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Actually, one second. I am solving it right now. Will post a solution.
 
  • #3
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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?
 
  • #4
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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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  • #5
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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.
 
  • #6
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Why do you put du=2xdx? I think we need dx=du/2x.

Here is the completed version.
 

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  • #7
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Can you check another problem please too? Thank you.
 

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  • #8
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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]
 
  • #9
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Thank you very much!

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

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  • #10
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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]

OH! This one is much prettier than mine!
 
  • #11
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Thank you very much!

I am also stuck on this one. ex has been always tricky for me.
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.
 
  • #12
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Thank you a lot, Mark44!
 

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