Integration by substitution

**sinewave85** · August 26th 2009, 08:15 AM

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

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

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

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

mathmatica gives an answer of

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

I tries using $u=\sqrt{1-x^{2}}$ , but that did not produce an equation I could integrate. Does anyone have any suggestions?

**chengbin** · August 26th 2009, 08:40 AM

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

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

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

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

**sinewave85** · August 26th 2009, 08:55 AM

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!

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