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.

  • 2
    $\begingroup$ Hint: $du\leftrightarrow dw$. $\endgroup$
    – user65203
    May 12, 2015 at 21:02
  • 2
    $\begingroup$ 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


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.

  • $\begingroup$ 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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.