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I have the following integral $$\int \frac{x}{\sqrt{4-x^2}}dx$$

So I do $u$ substitution $$u = -x^2 + 4$$ $$du = -2x \, dx$$

and get the following $$-\frac12\int \frac1{\sqrt{u}}du$$

I then can get TWO answers

1) Using $\int\frac1x = \ln(x)$

   integral[1/sqrt(u)]  ; the integral as it is to this point
   integral[1/w]        ; w = sqrt(u)
   ln(w)                ; evaluate integral
   ln(sqrt(u))          ; replace w
   ln(sqrt(4-x^2))      ; replace u

2) Use general power rule

   integral[u^(-1/2)]  ; Rewriting 1/sqrt(u) as u^(-1/2)
   u^(1/2) / (1/2)     ; evaluate using power rule
   (2/-2) * u^(1/2)    ; rewrite the (1/2) divisor as multiplying by 2
   -sqrt(u)            ; write u^(1/2) as sqrt(u)
   -sqrt(4-x^2)        ; replace u

What's going on? Is it that I can't make the replacement of sqrt(u) with a w? Why not? isn't it just a place holder, so to speak.

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    $\begingroup$ Hint: $du\leftrightarrow dw$. $\endgroup$
    – user65203
    May 12, 2015 at 21:02
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    $\begingroup$ The problem isn't with $w = \frac 1{\sqrt u}$. Rather, you need to account for $dw = -\frac 12 u^{-3/2} \,du$ $\endgroup$ May 12, 2015 at 21:03
  • $\begingroup$ So when you 'set up' w you have to also have dw in the equation? $\endgroup$
    – Zimm3r
    May 12, 2015 at 21:04
  • $\begingroup$ In your first substitution you accounted for $du$, why didn't you do the same in the second substitution? Also, you're missing a $dx$ in $du = -2x dx$. $\endgroup$
    – Andre
    May 12, 2015 at 21:10

3 Answers 3

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The problem isn't with $w = {\sqrt u}$. Rather, you failed to account for $$dw = \frac 12 u^{-1/2} \,du = \frac 12\cdot \frac {du}{\sqrt u}$$

When you chose to express $w$ as a function of $u$, you need also to express $dw$ in terms of $du$. If we do this, note that $$-1/2 \int \frac {du}{\sqrt u} = -\int dw = -w + C = -\sqrt u+C = -\sqrt{4-x^2} + C$$

The second method is the most straightforward way to go: using the power rule.

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  • $\begingroup$ I don't get why this was voted down. Nothing said here is wrong in my opinion. $\endgroup$
    – randomgirl
    May 12, 2015 at 21:09
  • $\begingroup$ I'm not clear about that, either. $\endgroup$ May 12, 2015 at 21:10
  • $\begingroup$ @JordanGlen I just posted an answer (a correct one) to another question and received a down vote even though the OP thanked me. There are some users that seem to take pleasure on punishing those who are sincerely trying to help. +1 for you. $\endgroup$
    – Mark Viola
    May 12, 2015 at 23:30
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The power rule is a valid way of solving the problem. If you decide to write $w=\sqrt{u}$, then you also need to make a substitution for $du$ based on $dw=\frac{1}{2\sqrt{u}}\,du$.

This is a good example of why you should avoid shorthand when writing integrals, i.e. write $$\int\frac{du}{\sqrt{u}}$$ rather than $$\int\frac{1}{\sqrt{u}}.$$

Edit: Note that the $dx$ and $du$ currently in the question were added by another user's edit.

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You forgot the dx. Where did it go? I don't think the integral involving the $u$ is OK. You need to express dx in terms of du before even getting to the u form of your integral.

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