yes it should. :( Check out the hundred other posts here, I apologize.

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- Jan 23rd 2010, 09:42 PMVonNemo19
- Jan 23rd 2010, 09:44 PMProve It
- Jan 24th 2010, 03:04 AMArchie Meade
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

$\displaystyle \int{\frac{1}{\sqrt{4-64x^2}}}dx=\int{\frac{1}{\sqrt{2^2-(8x)^2}}}dx=\int{\frac{1}{\sqrt{2^2-y^2}}}\frac{dy}{8}$

$\displaystyle =\frac{1}{8}\int{\frac{1}{\sqrt{2^2-y^2}}}dy=\frac{1}{8}arcsin(4x)+C$ - Jan 24th 2010, 07:26 AMnaomi
thank you all for your help.

- Jan 26th 2010, 12:46 PMnaomi
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))? - Jan 26th 2010, 07:29 PMProve It
$\displaystyle \sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\sin{\theta}\right)^2} = \sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2\sin^2{\theta}}$

$\displaystyle = \sqrt{\left(\frac{1}{4}\right)^2(1 - \sin^2{\theta})}$

$\displaystyle = \sqrt{\left(\frac{1}{4}\right)^2}\sqrt{1 - \sin^2{\theta}}$

$\displaystyle = \frac{1}{4}\sqrt{1 - \sin^2{\theta}}$

$\displaystyle = \frac{1}{4}\sqrt{\cos^2{\theta}}$

$\displaystyle = \frac{1}{4}\cos{\theta}$.

Step 1: Take the square through all the factors.

Step 2: Take out a common factor of $\displaystyle \left(\frac{1}{4}\right)^2$. - Jan 26th 2010, 11:27 PMnaomi
Hi Prove it

Could you explain how you get from

sqrt(-(1/4)^2((sin^2)(theta)))

to

sqrt((1-sin^2(theta))) - Jan 27th 2010, 08:34 PMProve It
Check my edit.

- Jan 28th 2010, 12:06 AMnaomi
Thank you.