# Math Help - integral

1. Originally Posted by naomi
@ VonNemo19:
yes it should. Check out the hundred other posts here, I apologize.

2. Originally Posted by VonNemo19
No, but I would like to hear what makes you ask that question.
Yes it would. You have forgotten to take into account that $\frac{du}{dx} = 8$.

So you will need to turn the integral into

$\frac{1}{8}\int{8f\left(u(x)\right)\,dx}$.

3. hi naomi,

following on from Prove It's good post,

in terms of the substitution you are working with,
we have

$\int{\frac{1}{\sqrt{4-64x^2}}}dx$

If we write this as

$\int{\frac{1}{\sqrt{a^2-y^2}}}dy$

then

$y^2=64x^2=(8x)^2$

$y=8x$

$\frac{dy}{dx}=8$

$dy=8dx$

$dx=\frac{dy}{8}$

hence

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

$=\frac{1}{8}\int{\frac{1}{\sqrt{2^2-y^2}}}dy=\frac{1}{8}arcsin(4x)+C$

4. thank you all for your help.

5. 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))?

6. Originally Posted by naomi
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))?
$\sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\sin{\theta}\right)^2} = \sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2\sin^2{\theta}}$

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

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

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

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

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

Step 1: Take the square through all the factors.

Step 2: Take out a common factor of $\left(\frac{1}{4}\right)^2$.

7. Hi Prove it

Could you explain how you get from

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

8. Check my edit.

9. Thank you.

Page 2 of 2 First 12