hi naomi,
following on from Prove It's good post,
in terms of the substitution you are working with,
we have
$\displaystyle \int{\frac{1}{\sqrt{4-64x^2}}}dx$
If we write this as
$\displaystyle \int{\frac{1}{\sqrt{a^2-y^2}}}dy$
then
$\displaystyle y^2=64x^2=(8x)^2$
$\displaystyle y=8x$
$\displaystyle \frac{dy}{dx}=8$
$\displaystyle dy=8dx$
$\displaystyle dx=\frac{dy}{8}$
hence
$\displaystyle \int{\frac{1}{\sqrt{4-64x^2}}}dx=\int{\frac{1}{\sqrt{2^2-(8x)^2}}}dx=\int{\frac{1}{\sqrt{2^2-y^2}}}\frac{dy}{8}$
$\displaystyle =\frac{1}{8}\int{\frac{1}{\sqrt{2^2-y^2}}}dy=\frac{1}{8}arcsin(4x)+C$
Hi Prove it,
your trigonometric approach is almost clear to me. There is one thing which I don't completely get.
How do you get from this step sqrt((1/4)^2-(1/4sin (theta))^2) to 1/4 sqrt(1-sin^2(theta))?
$\displaystyle \sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\sin{\theta}\right)^2} = \sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2\sin^2{\theta}}$
$\displaystyle = \sqrt{\left(\frac{1}{4}\right)^2(1 - \sin^2{\theta})}$
$\displaystyle = \sqrt{\left(\frac{1}{4}\right)^2}\sqrt{1 - \sin^2{\theta}}$
$\displaystyle = \frac{1}{4}\sqrt{1 - \sin^2{\theta}}$
$\displaystyle = \frac{1}{4}\sqrt{\cos^2{\theta}}$
$\displaystyle = \frac{1}{4}\cos{\theta}$.
Step 1: Take the square through all the factors.
Step 2: Take out a common factor of $\displaystyle \left(\frac{1}{4}\right)^2$.