why is the following U-Substitution wrong?

Question

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) $

However,

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.

REMARK

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

Example

If $ u = 1-x^2 $ then $d\color{red}u = -2\color{red}x d\color{red}x$ — J. W. Tanner, May 02 '19 at 05:18
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 $\sin^{-1}(x) = 1/(\sqrt{1-x^2})$. — D.B., May 02 '19 at 05:19

score 1 · Answer 1 · answered May 02 '19 at 05:23

1

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

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

answered May 02 '19 at 05:23

J. W. Tanner

1 Answers1