Hey,
Unfortunately I am not able to solve the following integral:
∫1/(SQRT(4-64(x)^2))dx
Can somebody help me?
Thanks
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
$\displaystyle \int{\frac{1}{\sqrt{4-64x^2}}}dx=\frac{1}{8}\int{\frac{1}{Cos\theta}}dSi n\theta$
@ 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.
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
$\displaystyle a^2=4,\ a=2$
$\displaystyle y^2=64x^2=(8x)^2,\ y=8x$
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 $\displaystyle \frac{1}{8}\arcsin{4x} + C$.