# Integration by substitution

• Aug 26th 2009, 08:15 AM
sinewave85
Integration by substitution
I was checking my work against the mathematica integral calculator, and my answer on this problem is totaly different than theirs. Can anyone tell me what I did wrong?

$\int\frac{xdx}{\sqrt{1-x^{2}}}$

I did

$\mbox{Let }u=1-x^{2}, du=-2xdx$

$\int\frac{xdx}{\sqrt{1-x^{2}}}=\frac{-1}{2}\int(\sqrt{u})^{-1}du$

$\int\frac{xdx}{\sqrt{1-x^{2}}}=\frac{-1}{2}\ln|\sqrt{1-x^{2}}|+c$

$-\sqrt{1-x^{2}}$

I tries using $u=\sqrt{1-x^{2}}$, but that did not produce an equation I could integrate. Does anyone have any suggestions?
• Aug 26th 2009, 08:40 AM
chengbin
You don't integrate $\frac {1}{\sqrt u}$ like that, you integrate it normally, so

$\int \frac {1}{\sqrt u}du$

$\int u^{-\frac {1}{2}}du$

$2u^{\frac {1}{2}}+c$
• Aug 26th 2009, 08:55 AM
sinewave85
Quote:

Originally Posted by chengbin
You don't integrate $\frac {1}{\sqrt u}$ like that, you integrate it normally, so

$\int \frac {1}{\sqrt u}du$

$\int u^{-\frac {1}{2}}du$

$2u^{\frac {1}{2}}+c$

Do'h. Thanks very much!