# More Inverse Trig.

• Nov 15th 2009, 08:53 AM
xwrathbringerx
More Inverse Trig.
Use the substitution $u = \arcsin (x/a)$to find the integral of $\sqrt(a^2-x^2)$ between x = 0 and x = b.

I've never confronted an integral as hard as this. Could someone please please show me how to do this???
• Nov 15th 2009, 09:16 AM
skeeter
Quote:

Originally Posted by xwrathbringerx
Use the substitution $u = \arcsin (x/a)$to find the integral of $\sqrt(a^2-x^2)$ between x = 0 and x = b.

I've never confronted an integral as hard as this. Could someone please please show me how to do this???

$\sin{u} = \frac{x}{a}$

$x = a\sin{u}$

$dx = a\cos{u} \, du$

$\int \sqrt{a^2 - x^2} \, dx$

$\int \sqrt{a^2 - a^2\sin^2{u}} \cdot a\cos{u} \, du$

$\int \sqrt{a^2(1 - \sin^2{u})} \cdot a\cos{u} \, du$

$a^2 \int \cos^2{u} \, du$

use the double angle identity $\cos^2{u} = \frac{1+\cos(2u)}{2}$ to help in finding the antiderivative.