# 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?

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

I did

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

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

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

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

I tries using $\displaystyle u=\sqrt{1-x^{2}}$, 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 $\displaystyle \frac {1}{\sqrt u}$ like that, you integrate it normally, so

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

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

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

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

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

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

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

Do'h. Thanks very much!