# integral

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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
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:

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

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
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:

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
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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