Integration by substitution

  • Thread starter flipsvibe
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  • #1
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Homework Statement


Compute the indefinite integral.

∫(x^2 + 1)^(-5/2) dx

The Attempt at a Solution


I have a hunch that I need to substitute x = tan(u) but, as always, my lack of trig skills are holding me back.
 

Answers and Replies

  • #2
35,642
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Yes, that's the right substition.

If x = tan u, dx = ?
 
  • #3
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So dx = sec^2 u du?
 
  • #4
35,642
7,517
Right. And you also have the du, so good job. You'll need to substitute for the x^2 + 1 part, so that's tan^2(x) + 1, right?

When you're done with you trig substitution, your new integral will be completely in terms of u and du, with x and dx completely gone.
 
  • #5
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I plugged in the x = tan u and dx = sec^2 u du, and I got
∫[tan^2(u) + 1]^(-5/2) * sec^2(u) du
==> ∫[sec^2(u)]^(-5/2) * sec^2(u) du
==> ∫[sec^2(u)]^(-3/2) du
after integration I end up with (-3/2)(tanu/secu)
If I continue on and cancel out the cosines, I end up with (-3/2)sinu.
I don't know how to get this back in terms of x.
 
  • #6
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Wait, I think I got this.
I didn't write out all of my steps in my last reply..so I'll go back one or two, to one I skipped.
(-3/2)[tan(u) / √sec^2(u)]
==>(-3/2) [tanu / √(tan^2u + 1)]
==>-3x / 2√(x^2 + 1)
 
Last edited:
  • #7
35,642
7,517
Don't use ==> when you mean =.

Also, don't forget your constant of integration. I didn't check your answer, but you can do that. If your antiderivative is correct, you should be able to differentiate it and get back the original integrand. IOW, d/dx(-3x/2sqrt(x^2 + 1)) should equal (x^2 + 1)^(-5/2).
 

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