It is known that

$$ \int \frac{1}{\sqrt{1-x^2}} dx = arcsin(x)+c$$

this can be done utilizing u-substitution $ x = sin(u) $


i can let $ u = 1-x^2 $

$dx = -2u \, du $

which gives the integral

$$ -\int \frac{2u}{\sqrt{u}} du $$

which then returns $$ \frac{-4\sqrt{u^3}}{3} $$

which is clearly wrong.

So why is this wrong?

thanks very much for your help and apologies about the elementary question.


Upon checking other posts , a common problem has to do with the range of the function being substituted in.


What is wrong with the following u-substitution?

  • 1
    $\begingroup$ If $ u = 1-x^2 $ then $d\color{red}u = -2\color{red}x d\color{red}x$ $\endgroup$ – J. W. Tanner May 2 at 5:18
  • $\begingroup$ Your differentials are incorrect. With your u-sub $du = -2xdx$. This is not the same as $dx = -2udu$. I don't think that a u-sub is possible here. This is just a well-known integral, which is known based on the fact that the derivative of $\sin^{-1}(x) = 1/(\sqrt{1-x^2})$. $\endgroup$ – D.B. May 2 at 5:19

Your alternative method is incorrect, because if $u=1-x^2$ then $du=-2x\,dx$.

You wrote $dx=-2u\,du$; that is different.


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