Partial fractions integration problem

[Trenthan]

Homework Statement

Ey guys I am trying to derive a potential flow equation and in the process i need to integrate with respect to 'x'.

I need to integrate what is shown below with respect to x

Homework Equations

-\frac{y^2}{(x^{2} + y^{2})^{2}}

The Attempt at a Solution

okay I've run out of idea's
-tried many different substitutions
-tried partial fractions
-tried partial fractions and than make a substitution

im assuming it can be done analytically since the integral is in a nice form, from matlab
-atan(x/(y^2)^(1/2))/(2*(y^2)^(1/2)) - x/(2*(x^2 + y^2))

 

[issacnewton]

hi

express numerator as y^2=(x^2+y^2-x^2) then you can separate the integral
in two separate parts. first part is

\int \frac{dx}{(x^2+y^2)}

which is straight forward... hint... think of arc(tan) function...

second part is

- \int \frac{x^2 \,dx}{(x^2+y^2)^2}

make the substitution

x=y\tan(\theta)

this will simplify it...

 

[Trenthan]

make the substitution

x=y\tan(\theta)

this will simplify it...[/QUOTE]

Wouldn't it be x = y/tan(theta)

?

If not where did that relationship come from?

 

[issacnewton]

this is standard substitution in these types of problems... here the variable is x and y is
constant..

 

[Trenthan]

x=y\tan(\theta)

havent done these for quite some time going through the year 12 textbook atvm hoping we covered it than. If not uni notes next

Quick question we take the diff of above with respect to theta ot y?

ive made the substitution etc and now have this ugly thing in terms of tan's and sec's

 

[issacnewton]

first part is (refer to

[PLAIN]http://www.sosmath.com/tables/integral/integ8/integ8.html

[/PLAIN]

even the first part can be done with the substitution suggested by me. )

<br /> \int \frac{dx}{(x^2+y^2)} = \frac{1}{y}\tan^{-1}\frac{x}{y}<br />

now coming to the second part ,let's call the integral , I.

I=- \int \frac{x^2 \,dx}{(x^2+y^2)^2}

let <br /> x=y\tan(\theta)<br />

where \theta is the new variable. so

dx = y \sec^{2}\theta \,d\theta

so I becomes

I=-\int \frac{y^2(\tan^{2}\theta)y\sec^{2}\theta\,d\theta}{(y^2\tan^{2}\theta+y^2)^2}

can you simplify it ?

 

[Trenthan]

bit of playing around got me to there,

with abit more i got down to this, (see attached)


the first arc tan looks good, when combined with the other arc tan which u found, (i just ignored that bit for now) i get the "atan(x/(y^2)^(1/2))/(2*(y^2)^(1/2)) " from matlab


its the 2nd term "- x/(2*(x^2 + y^2))", I am not sure how to transform the sin(2*theta)

 

[issacnewton]

you are almost right. the very last term at the bottom should be

\frac{1}{4y}\sin \left[2\tan^{-1}\left(\frac{x}{y}\right)\right]

 

[Trenthan]

you are almost right. the very last term at the bottom should be

\frac{1}{4y}\sin \left[2\tan^{-1}\left(\frac{x}{y}\right)\right]

how would i transform that into

x/(2*(x^2 + y^2))

since that's what I am looking for :S

well where would i start in order to do that, mat-lab was right matches no.10 off the site u posted :)

 

[issacnewton]

so the whole integral is

\frac{1}{y}\tan^{-1}\frac{x}{y}-\frac{1}{2y}\tan^{-1}\left(\frac{x}{y}\right)+\frac{1}{4y}\sin \left[2\tan^{-1}\left(\frac{x}{y}\right)\right]

 

[Trenthan]

btw TY IssacNewton ! huge help

so the whole integral is

\frac{1}{y}\tan^{-1}\frac{x}{y}-\frac{1}{2y}\tan^{-1}\left(\frac{x}{y}\right)+\frac{1}{4y}\sin \left[2\tan^{-1}\left(\frac{x}{y}\right)\right]

yep that's it, how would i simply it down to
atan(x/(y^2)^(1/2))/(2*(y^2)^(1/2)) - x/(2*(x^2 + y^2))
?

The tan bit i can easily do but the last term...

(Only asking since to complete the remainder which i use this to create my complex velocity potential, the sins and tans won't allow me to simply it down for the remainder of the questions. They have used "- x/(2*(x^2 + y^2))" during this process)

 

[issacnewton]

the first two terms become

\frac{1}{2y}\tan^{-1}\left(\frac{x}{y}\right)

now look at the last term

\frac{1}{4y}\sin \left[2\tan^{-1}\left(\frac{x}{y}\right)\right]

we can let
\tan^{-1}\left(\frac{x}{y}\right) =\alpha

then

\tan \alpha=\frac{x}{y}

from here we can get

\sin \alpha and

\cos \alpha

which are nothing but

\sin\left[\tan^{-1}\left(\frac{x}{y}\right)\right]

and

\cos\left[\tan^{-1}\left(\frac{x}{y}\right)\right]

now using trig identities you can find

\sin \alpha

and

\cos \alpha

 

[issacnewton]

so the last term<br /> \frac{1}{4y}\sin \left[2\tan^{-1}\left(\frac{x}{y}\right)\right]<br />

is

\frac{1}{4y}\sin(2\alpha)

which is

\frac{1}{4y}\cdot 2\sin \alpha \cos \alpha

you already know that \tan \alpha =\frac{x}{y}

use trig identities to find
\sin \alpha

and

\cos \alpha in terms of x and y

and I saw that your integral had a minus sign in the first post. I didn't see it. So take
that into account too..