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January 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
January 23rd 2010, 12:53 PM
pickslides

Use $\int \frac{1}{\sqrt{1-u^2}}~du = \sin^{-1}u+C$
January 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.
January 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.

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

Therefore use the sub $u=4x$
January 23rd 2010, 01:02 PM
Archie Meade

Use

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

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

Quote:

Originally Posted by Archie Meade

Use

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

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

$\sqrt{4-64x^2}= \sqrt{4}\sqrt{1-16x^2}=2\sqrt{1-(4x)^2} \neq\sqrt{2}\sqrt{1-16x^2}$
January 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 $\int\frac{1}{\sqrt{a^2-u^2}}du=\arcsin\frac{u}{a}+C$ .

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

Quote:

Originally Posted by VonNemo19

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

I read the original post to be

$\int\frac{1}{\sqrt{4-64x^2}}dx$
January 23rd 2010, 01:30 PM
naomi

u=4x, du=4dx
∫1/(2sqrt(1-(u)^2))dx
January 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?
January 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

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

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

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

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

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

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

hence

$\int{\frac{1}{\sqrt{4-64x^2}}}dx=\frac{1}{8}\int{\frac{1}{Cos\theta}}dSi n\theta$
January 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, $a=2$ and $u=8x$ so $\int\frac{1}{\sqrt{(2)^2-(8x)^2}}dx=\arcsin\frac{8x}{2}+C=\arcsin(4x)+C$
January 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.
January 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.

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

dx needs to be changed to dy

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

$y^2=64x^2=(8x)^2,\ y=8x$
January 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 $\int{\frac{1}{\sqrt{4 - 64x^2}}\,dx} = \int{\frac{1}{\sqrt{64\left(\frac{1}{16} - x^2\right)}}\,dx}$

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

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

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

So the integral becomes

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

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

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

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

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

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

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

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

$4x = \sin{\theta}$

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

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

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