1. ## Integral question

I'm sorry in advance that I don't know how to make this look nice like the other posts do.

I have this integral $xdy/(x^2 + y^2)^(3/2) where x is a constant. The answer is given to me as y/x(x^2 + y^2)^(1/2) I am able to work backwards by taking the derivative of the answer to get the original integral but I can't figure out how to work the integral. Any help would be appreciated, Thank you 2. Originally Posted by Quasar I'm sorry in advance that I don't know how to make this look nice like the other posts do. I have this integral$ xdy/(x^2 + y^2)^(3/2) where x is a constant.

The answer is given to me as y/x(x^2 + y^2)^(1/2)

I am able to work backwards by taking the derivative of the answer to get the original integral but I can't figure out how to work the integral.

Any help would be appreciated,

Thank you
they used a trigonometric substitution, $y = x \tan \theta$

(remember, we treat x as a constant throughout the entire problem)

can you do the integral now that you know that?

to make the math symbols "look nice," we use LaTex

3. I still can't do the integral but atleast I know where to start looking. I have my math book open to the Trigonometric Substitutions chapter now. I'm taking Physics II and Calc II this semester and this is actually part of a physics problem I have involving electric fields. It looks like I'm 2 to 3 weeks away from learning the Trig Substitution method in my Calc II class. I worked through this physics problem and I got the integral set up but then I couldn't figure out what to do with the part of it that I just posted.

I've been going around and around with this problem now for hours. I thought that I must know how to work that integral but I just couldn't figure it out. I have a test in the morning so I need to get this figured out tonight...if I don't pass out first. I'm exhausted.

4. Originally Posted by Quasar
I still can't do the integral but atleast I know where to start looking. I have my math book open to the Trigonometric Substitutions chapter now. I'm taking Physics II and Calc II this semester and this is actually part of a physics problem I have involving electric fields. It looks like I'm 2 to 3 weeks away from learning the Trig Substitution method in my Calc II class. I worked through this physics problem and I got the integral set up but then I couldn't figure out what to do with the part of it that I just posted.

I've been going around and around with this problem now for hours. I thought that I must know how to work that integral but I just couldn't figure it out. I have a test in the morning so I need to get this figured out tonight...if I don't pass out first. I'm exhausted.

if you haven't done it in your class, why are you expected to know it for the test?

$\int \frac {x~dy}{\left( x^2 + y^2 \right)^{3/2}}$

Let $y = x \tan \theta \implies \tan \theta = \frac yx$ ....remember this, we will get back to it

$\Rightarrow dy = x \sec^2 \theta ~d \theta$

So our integral becomes:

$\int \frac {x^2 \sec^2 \theta}{\left[ x^2 \left( 1 + \tan^2 \theta \right) \right]^{3/2}}~d \theta = \frac 1x \int \frac {\sec^2 \theta}{\sec^3 \theta}~d \theta$

$= \frac 1x \int \frac 1{\sec \theta}~d \theta$

$= \frac 1x \int \cos \theta ~d \theta$

$= \frac 1x \sin \theta + C$

now we have an answer in terms of $\theta$, but we want an answer in terms of y

now recall that we said $\tan \theta = \frac yx$

so if we draw a right triangle with an acute angle $\theta$, the side opposite to the angle will be $y$ and the side adjacent the angle will be $x$. By Pythagoras, the hypotenuse will be $\sqrt{x^2 + y^2}$.

Now $\sin \theta = \frac {\mbox {Opposite}}{\mbox {Hypotenuse}} = \frac y{\sqrt {x^2 + y^2}}$

plugging this in for $\sin \theta$ in our answer, we find that:

$\int \frac {x~dy}{\left( x^2 + y^2 \right)^{3/2}} = \frac y{x \sqrt {x^2 + y^2}} + C$

as desired

5. This integral's getting interesting, look at here.