`int u sqrt(1 - u^2) du` Evaluate the indefinite integral.

Friday, October 9, 2009

$\displaystyle\int{u}\sqrt{{{1}-{{u}}^{{2}}}}{d}{u}$ Evaluate the indefinite integral.

You need to
use the following substitution $\displaystyle{1}-{{u}}^{{2}}={t}$ , such that:

$\displaystyle{1}-{{u}}^{{2}}={t}\Rightarrow$
-2udu = dt=>u du = -(dt)/2

$\displaystyle\int{u}\cdot\sqrt{{{1}-{{u}}^{{2}}}}{d}{u}=-{\left(\frac{{1}}{{2}}\right)}\cdot\int\sqrt{{t}}$
dt

$\displaystyle-{\left(\frac{{1}}{{2}}\right)}\cdot\int\sqrt{{t}}{\left.{d}{t}\right.}={\left(-\frac{{1}}{{2}}\right)}\cdot\frac{{{{t}}^{{\frac{{3}}{{2}}}}}}{{\frac{{3}}{{2}}}}+{c}$

Replacing back $\displaystyle{1}-{{u}}^{{2}}$ for t yields:

$\displaystyle\int{u}\cdot\sqrt{{{1}-{{u}}^{{2}}}}{d}{u}=$
(-1/3)*((1 - u^2)^(3/2)) + c

Hence, evaluating the indefinite
integral, yields $\displaystyle\int{u}\cdot\sqrt{{{1}-{{u}}^{{2}}}}{d}{u}=-\frac{{{{\left({1}-{{u}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}}{{3}}+{c}$

Spooky Notes

Friday, October 9, 2009

$\displaystyle\int{u}\sqrt{{{1}-{{u}}^{{2}}}}{d}{u}$ Evaluate the indefinite integral.

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