# Math Help - integral

1. ## integral

Hey,

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

Can somebody help me?
Thanks

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

3. 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.

4. 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 $u=4x$

5. Use

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

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

6. Originally Posted by Archie Meade
Use

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

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

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

7. 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 $u=8x$.

8. Originally Posted by VonNemo19
$\int\sqrt{4-64x^2}dx$
I read the original post to be

$\int\frac{1}{\sqrt{4-64x^2}}dx$

9. u=4x, du=4dx
∫1/(2sqrt(1-(u)^2))dx

10. 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?

11. 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

$\int{\frac{1}{\sqrt{4-64x^2}}}dx=\frac{1}{8}\int{\frac{1}{Cos\theta}}dSi n\theta$

12. 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 $\int\frac{1}{\sqrt{(2)^2-(8x)^2}}dx=\arcsin\frac{8x}{2}+C=\arcsin(4x)+C$

13. @ 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.

14. 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

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

$y^2=64x^2=(8x)^2,\ y=8x$

15. 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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