Question

Integrate sqrt(1-64x^2)?
Medal for help! :)

I know it's a sub with Sine, but I don't know how to use it here. please help

Answer 1

\[\Large\rm \int\limits \sqrt{1-64x^2}~dx\]Bring the 64 into the square,\[\Large\rm \int\limits\limits \sqrt{1-(8x)^2}~dx\]Understand what I did there?
That will make our substitution a little easier.
We have the form \(\Large\rm 1-(stuff)^2\) and we make the substitution \(\Large\rm stuff=\sin\theta\)

Answer 2

then what is x? I can't see a way to clear the radical

Answer 3

x = sin(t)/8?

Answer 4

So our substitution is:\[\Large\rm 8x=\sin t\]
This is what we're subbing into our integral.
We're replacing BOTH the 8 and the x with sin t

Answer 5

\[\Large\rm \int\limits\limits\limits \sqrt{1-(\color{royalblue}{8x})^2}~dx\]\[\Large\rm \int\limits\limits\limits \sqrt{1-(\color{royalblue}{\sin t})^2}~dx\]

Answer 6

oh I see

Answer 7

This is in that really nice beautiful form now!
It allows us to use our Pythagorean Identity.

Answer 8

The whole reason we switch things into trig and do all of this messy stuff,
is to get rid of the addition or subtraction between the terms.

It's so much easier to take a root when we have a single term.

Answer 9

Remember your Pythagorean Identity for Sine and Cosine? :)

Answer 10

yes, I know about the triangle reference. so the y in sine would be 8x, right?

Answer 11

Ummm yes very good.
And the hypotenuse 1 in that case.

Answer 12

ok I got it from here. thanks for your help! :)

Answer 13

Cool :)

Answer 14

Don't forget you have to substitute for your differential `dx` as well!
Just reminding you hehe