Integral dx/sqrt(x - a)integral dx/sqrt(1/ax)

  • Thread starter rebeka
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  • #1
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integral dx/sqrt(x - a)
integral dx/sqrt(1/ax)
 
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  • #2
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reading suggestions
 
  • #3
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sorry for wasting time I think I just understood substitution, out of embarassment im never coming back to this forum(bright side of life I learned more in three days than I have in two years)
 
  • #4
HallsofIvy
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We all embarrass ourselves at one time or another! (Some of us more than others. Believe me, I know about embarrassing myself!) PLEASE come back to PhysicsForums!

(I need someone to make me look good!!!)

Anyway, for those who are still wondering how to do these:

[tex]\int \sqrt{x-a}dx[/tex]

Let u= x-1 so du= dx and [tex]\sqrt{x-a}= \sqrt{u}= u^{\frac{1}{2}}[/tex]. The integral becomes [tex]\int u^{\frac{1}{2}}du[/tex] which can be done with the power law.

[tex]\int \frac{dx}{\sqrt{\frac{1}{ax}}}[/tex]

[tex]\frac{1}{\sqrt{\frac{1}{ax}}}[/tex] is just [tex]\sqrt{ax}[/tex].

Let u= ax so du= adx, dx= (1/a)du and [tex]\sqrt{ax}= \sqrt{u}= u^{\frac{1}{2}}[/tex]

The integral becomes [tex]\frac{1}{a}\int u^{\frac{1}{2}}du[/tex]
 
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  • #5
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HallsofIvy said:
[tex]\int \sqrt{x-a}dx[/tex]

HallsofIvy, should not this equation be:
[tex]\int \frac{dx}{\sqrt{x - a}} = \int u^{-\frac{1}{2}} du[/tex]
 
  • #6
HallsofIvy
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Oh, you're right! I missed the "1/" in the first post. Of course, the substitution would be exactly the same and the u-integral what you show. Thanks.
 

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