# integral

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• Jan 23rd 2010, 12:40 PM
naomi
integral
Hey,

Unfortunately I am not able to solve the following integral:
∫1/(SQRT(4-64(x)^2))dx

Can somebody help me?
Thanks
• Jan 23rd 2010, 12:53 PM
pickslides
Use $\displaystyle \int \frac{1}{\sqrt{1-u^2}}~du = \sin^{-1}u+C$
• Jan 23rd 2010, 12:58 PM
naomi
Thanks for your reply. I had that formula in mind as well, but don't really have an idea how to apply it to this exercise.
• Jan 23rd 2010, 01:00 PM
e^(i*pi)
Quote:

Originally Posted by naomi
Thanks for your reply. I had that formula in mind as well, but don't not really have an idea how to apply it to this exercise.

$\displaystyle 4-64x^2 = 4(1-16x^2)$

Therefore use the sub $\displaystyle u=4x$
• Jan 23rd 2010, 01:02 PM
Archie Meade
Use

$\displaystyle 4-64x^2=4(1-16x^2)$

$\displaystyle \sqrt{4-64x^2}=\sqrt{4}\sqrt{1-16x^2}=\sqrt{4}\sqrt{1-(4x)^2}$
• Jan 23rd 2010, 01:07 PM
pickslides
Quote:

Originally Posted by Archie Meade
Use

$\displaystyle 4-64x^2=4(1-16x^2)$

$\displaystyle \sqrt{4-64x^2}= \sqrt{2}\sqrt{1-16x^2}=\sqrt{2}\sqrt{1-(4x)^2}$

$\displaystyle \sqrt{4-64x^2}= \sqrt{4}\sqrt{1-16x^2}=2\sqrt{1-(4x)^2} \neq\sqrt{2}\sqrt{1-16x^2}$
• Jan 23rd 2010, 01:24 PM
VonNemo19
Quote:

Originally Posted by naomi
Hey,

Unfortunately I am not able to solve the following integral:
∫1/(SQRT(4-64(x)^2))dx

Can somebody help me?
Thanks

EDIT: I missread the post

What we have is of the form $\displaystyle \int\frac{1}{\sqrt{a^2-u^2}}du=\arcsin\frac{u}{a}+C$ .

In this case, it is easy to see that $\displaystyle a=2$ and $\displaystyle u=8x$.
• Jan 23rd 2010, 01:28 PM
pickslides
Quote:

Originally Posted by VonNemo19
$\displaystyle \int\sqrt{4-64x^2}dx$

I read the original post to be

$\displaystyle \int\frac{1}{\sqrt{4-64x^2}}dx$
• Jan 23rd 2010, 01:30 PM
naomi
u=4x, du=4dx
∫1/(2sqrt(1-(u)^2))dx
• Jan 23rd 2010, 01:46 PM
naomi
so when a=2 and u = 8x then,
∫1/(sqrt(2^2-(8x)^2))dx
How can I put this into the following format: arcsin u/a +c?
• Jan 23rd 2010, 01:50 PM
Archie Meade
Hi naomi,

you can also use trigonometric substitutions.

One way is to draw a right-angled triangle with
perpendicular sides

$\displaystyle \sqrt{4-64x^2}\$ and $\displaystyle 8x$.

The hypotenuse of this triangle is 2, since $\displaystyle 2^2-(8x)^2=4-64x^2$

Then, with the acute angle opposite the side of length 8x, we get

$\displaystyle Sin\theta=\frac{8x}{2}=4x$

$\displaystyle \frac{d}{dx}Sin\theta=4$

$\displaystyle dx=\frac{1}{4}dSin\theta$

hence

$\displaystyle \int{\frac{1}{\sqrt{4-64x^2}}}dx=\frac{1}{8}\int{\frac{1}{Cos\theta}}dSi n\theta$
• Jan 23rd 2010, 01:51 PM
VonNemo19
Quote:

Originally Posted by naomi
so when a=2 and u = 8x then,
∫1/(sqrt(2^2-(8x)^2))dx
How can I put this into the following format: arcsin u/a +c?

Well, $\displaystyle a=2$ and $\displaystyle u=8x$ so $\displaystyle \int\frac{1}{\sqrt{(2)^2-(8x)^2}}dx=\arcsin\frac{8x}{2}+C=\arcsin(4x)+C$
• Jan 23rd 2010, 02:16 PM
naomi
@ VonNemo19:
Shouldn't the answer be: 1/8 arcsin(4x)+c instead of arcsin(4x)+c?

@ archie meade

Thanks for your explanation about the application of trigonometric substitutions to this exercise. I didn't know I could use this for this exercise. Trigonometric substitutions are quite new to me.
• Jan 23rd 2010, 02:35 PM
Archie Meade
Hi naomi,

yes, the trig substitutions can be tried later in your development.

It's best for you to master the inverse sine solution
recommended by the other contributors earlier.

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

dx needs to be changed to dy

$\displaystyle a^2=4,\ a=2$

$\displaystyle y^2=64x^2=(8x)^2,\ y=8x$
• Jan 23rd 2010, 09:28 PM
Prove It
Quote:

Originally Posted by naomi
Hey,

Unfortunately I am not able to solve the following integral:
∫1/(SQRT(4-64(x)^2))dx

Can somebody help me?
Thanks

I would use trigonometric substitution in this case.

You have $\displaystyle \int{\frac{1}{\sqrt{4 - 64x^2}}\,dx} = \int{\frac{1}{\sqrt{64\left(\frac{1}{16} - x^2\right)}}\,dx}$

$\displaystyle = \int{\frac{1}{8\sqrt{\frac{1}{16} - x^2}}\,dx}$

$\displaystyle = \frac{1}{8}\int{\frac{1}{\sqrt{\left(\frac{1}{4}\r ight)^2 - x^2}}\,dx}$.

Now make the substitution $\displaystyle x = \frac{1}{4}\sin{\theta}$ so that $\displaystyle dx = \frac{1}{4}\cos{\theta}\,d\theta$.

So the integral becomes

$\displaystyle \frac{1}{8}\int{\frac{1}{\sqrt{\left(\frac{1}{4}\r ight)^2 - \left(\frac{1}{4}\sin{\theta}\right)^2}}\,\frac{1} {4}\cos{\theta}\,d\theta}$

$\displaystyle = \frac{1}{32}\int{\frac{1}{\frac{1}{4}\sqrt{1 - \sin^2{\theta}}}\,\cos{\theta}\,d\theta}$

$\displaystyle = \frac{1}{8}\int{\frac{\cos{\theta}}{\sqrt{1 - \sin^2{\theta}}}\,d\theta}$

$\displaystyle = \frac{1}{8}\int{\frac{\cos{\theta}}{\sqrt{\cos^2{\ theta}}}\,d\theta}$

$\displaystyle = \frac{1}{8}\int{\frac{\cos{\theta}}{\cos{\theta}}\ ,d\theta}$

$\displaystyle = \frac{1}{8}\int{1\,d\theta}$

$\displaystyle = \frac{1}{8}\theta + C$.

Now, remembering that you made the original substitution $\displaystyle x = \frac{1}{4}\sin{\theta}$

$\displaystyle 4x = \sin{\theta}$

$\displaystyle \theta = \arcsin{4x}$.

So your final answer is $\displaystyle \frac{1}{8}\arcsin{4x} + C$.
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