# A Hard Calculus Problem

• Jul 5th 2009, 03:42 PM
iPod
A Hard Calculus Problem
The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and
b^2<c. Show that

dx/dt=(t-x)/(t+b)

and hence integrate

1/√(x^2+2bx+c)

Verify by direct integration that your result holds also in the case b^2 = c if x+b > 0 but that
your result does not hold in the case b^2 = c if x + b < 0 .

[and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]
• Jul 5th 2009, 03:59 PM
VonNemo19
Quote:

Originally Posted by iPod
The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and

b^2<c. Show that

dx/dt=(t-x)/(t+b)

and hence integrate

1/√(x^2+2bx+c)

Verify by direct integration that your result holds also in the case b^2 = c if x+b > 0 but that
your result does not hold in the case b^2 = c if x + b < 0 .

[and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]

Differentiate implicitly x with respect to t. This thread does not belong in this sub-forum.
• Jul 5th 2009, 09:42 PM
iPod
I managed to differentiate, however Im just stuck at the intergration part,
and doesnt this thread belong here??
• Jul 5th 2009, 10:36 PM
Hello iPod

Welcome to Math Help Forum!
Quote:

Originally Posted by iPod
The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and
b^2<c. Show that

dx/dt=(t-x)/(t+b)

[CENTER]
$\displaystyle t = x+\sqrt{x^2+2bx+c}$

$\displaystyle \Rightarrow (t-x)^2 = x^2+2bx+c$

$\displaystyle \Rightarrow 2(t-x)\left(\frac{dt}{dx}-1\right)=2x+2b$

$\displaystyle \Rightarrow \frac{dt}{dx} = 1+\frac{x+b}{t-x}=\frac{t+b}{t-x}$

$\displaystyle \Rightarrow \frac{dx}{dt} =\frac{t-x}{t+b}$

Quote:

and hence integrate

1/√(x^2+2bx+c)
$\displaystyle t-x = \sqrt{x^2+2bx+c}$

$\displaystyle \Rightarrow \frac{dx}{dt}=\frac{\sqrt{x^2+2bx+c}}{t+b}$

$\displaystyle \Rightarrow \int \frac{1}{\sqrt{x^2+2bx+c}}\,dx=\int\frac{1}{(t+b)} \,dt$

$\displaystyle = \ln (t+b)+C$

$\displaystyle =\ln(b+x+\sqrt{x^2+2bx+c})+C$, provided $\displaystyle b+x+\sqrt{x^2+2bx+c}>0$

Can you finish up?

PS
Quote:

[and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]
The language is called LaTeX, and you'll find some help on this site here: http://www.mathhelpforum.com/math-help/latex-help/

I also find the Wikipedia site helpful, here: LaTeX/Mathematics - Wikibooks, collection of open-content textbooks

When you've written some LaTex code, select what you've written and click the $\displaystyle \Sigma$ button on this editor's toolbar, and the editor will insert $$...$$ tags around the selection. Use 'Preview Changes' to check that it's OK.

To see any of the code I've used in my solution above, just click-left on any line and the code will appear in a separate pop-up window. Go on, try it now!

• Jul 6th 2009, 02:55 AM
simplependulum
Oh yeah , it is known as Euler's substitution for integration .

We can solve integrals in this form $\displaystyle \int \frac{ P(x) }{\sqrt{x^2 + 2ax + b}}~dx ~~$ using Euler's substitution $\displaystyle t = x + \sqrt{x^2 + ax + b}$

$\displaystyle dt = ( 1 + \frac{x + a }{\sqrt{x^2 + 2ax + b}})~dx$

$\displaystyle dt = [ \frac{ ( \sqrt{x^2 + 2ax + b}+x) + a}{ \sqrt{x^2 + 2ax + b}}]~dx$

Therefore , the integral becomes

$\displaystyle \int \frac{ P( f(t)) }{ (t+a)}~dx~dt$

Then , what does f(t) refer to ?

Since

$\displaystyle t = x + \sqrt{x^2 + 2ax + b}$
$\displaystyle t^2 - 2tx = 2ax + b$

$\displaystyle x = \frac{t^2 -b}{2(a+t)}$

$\displaystyle x = f(t) = \frac{t^2 -b}{2(a+t)}$

For more details , you can take a look at http://planetmath.org/encyclopedia/E...tegration.html .

Euler's substitution is a powerful tools indeed . Make use of it !!
• Jul 6th 2009, 02:12 PM
iPod
thx
Thank you very much Grandad and Simple, however I am not very familiar with this type of integration - is there a certain name for this type of integration and also is there a step by step method for it??
Thank you for your time,