Integrals containing $ \sqrt {a^2 - u^2}$ .

$\displaystyle \int \sqrt{a^2- u^2} \; du$ $\displaystyle = \arcsin- \frac ua + C.$ (B..24)
$\displaystyle \int \sqrt{a^2- u^2} \; du$ $\displaystyle = \frac u2 \sqrt{a^2-u^2} + \frac{a^2}{2} \arcsin \frac ua + C.$ (B..25)
$\displaystyle \int u^2 \; \sqrt{a^2-u^2} \; du$ $\displaystyle = \frac u8 (2u^2 - a^2 )\; \sqrt{a^2-u^2} + \frac {a^4}{8} \arcsin \frac ua + C.$ (B..26)
$\displaystyle \int \frac {\sqrt{a^2-u^2}}{ u} \; du$ $\displaystyle = \sqrt{a^2-u^2} - a \ln \begin{vmatrix}\frac {a + \sqrt{a^2-u^2}}{u} \end{vmatrix} + C.$ (B..27)
$\displaystyle \int \frac{\sqrt{a^2-u^2}}{u^2} \; du$ $\displaystyle = -\frac {\sqrt{a^2-u^2}}{u} - a \arcsin \frac ua + C.$ (B..28)
$\displaystyle \int \frac {u^2}{\sqrt{a^2-u^2}} \; du$ $\displaystyle = -\frac u2 \; \sqrt{a^2-u^2} + \frac {a^2}{2}\; \arcsin \frac ua + C.$ (B..29)
$\displaystyle \int \frac{du}{u \; \sqrt{a^2-u^2}}$ $\displaystyle = -\frac 1a \ln \begin{vmatrix}\frac{a + \sqrt{a^2-u^2}}{u} \end{vmatrix}+ C.$ (B..30)
$\displaystyle \int \frac{du}{u^2\; \sqrt{a^2-u^2}}$ $\displaystyle = - \frac{\sqrt{a^2-u^2}}{a^2 u} + C.$ (B..31)
$\displaystyle \int (a^2-u^2 )^{ 3/2}\; du$ $\displaystyle = -\frac u8 (2u^2- 5a^2 ) \; \sqrt{a^2-u^2} + \frac {3a^4}{8} \arcsin \frac ua + C.$ (B..32)
$\displaystyle \int \frac{du}{(a^2-u^2 )^{ 3/2}}$ $\displaystyle = \frac{u}{a^2 \; \sqrt{a^2-u^2}} + C.$ (B..33)

Noah Dana-Picard 2007-12-28