integral

**kasiaogor** · November 7th 2010, 02:02 PM

what is the integral of $1/\sqrt (36-16x^2)$

use substitution

**TheEmptySet** · November 7th 2010, 02:21 PM

Originally Posted by kasiaogor

what is the integral of $1/\sqrt (36-16x^2)$

use substitution

$\displaystyle \frac{1}{\sqrt{36-16x^2}}=\frac{1}{4}\frac{1}{\sqrt{\left( \frac{6}{4}\right)^2-x^2}}= \frac{1}{4}\frac{1}{\sqrt{\left( \frac{3}{2}\right)^2-x^2}}$

So the question you need to ask yourself is what trig sub works for things of the form

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

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