Hey,

Unfortunately I am not able to solve the following integral:

∫1/(SQRT(4-64(x)^2))dx

Can somebody help me?

Thanks

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- Jan 23rd 2010, 12:40 PMnaomiintegral
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 PMpickslides
Use $\displaystyle \int \frac{1}{\sqrt{1-u^2}}~du = \sin^{-1}u+C$

- Jan 23rd 2010, 12:58 PMnaomi
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 PMe^(i*pi)
- Jan 23rd 2010, 01:02 PMArchie 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 PMpickslides
- Jan 23rd 2010, 01:24 PMVonNemo19
- Jan 23rd 2010, 01:28 PMpickslides
- Jan 23rd 2010, 01:30 PMnaomi
u=4x, du=4dx

∫1/(2sqrt(1-(u)^2))dx - Jan 23rd 2010, 01:46 PMnaomi
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 PMArchie 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 PMVonNemo19
- Jan 23rd 2010, 02:16 PMnaomi
@ 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 PMArchie 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 PMProve It
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 $\displaystyle \frac{1}{8}\arcsin{4x} + C$.