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Two Integrals with Square Root

Barioth

Member
Jan 17, 2013
52
Hi everyone I have these 2 integrate that I can't solve, I have tried them with mathematica and wolfram, but they can't find an answer, maybe someone have an idea on how I could tackle these 2 bad boy!

The first one is

\(\displaystyle \int{ \sqrt{ \frac{1+( \frac{1}{10}+ \frac{s}{25})^2}{ \frac {s}{10}+ \frac{s^2}{50}}}}ds \)

The second one is

\(\displaystyle \int{ \sqrt{ \frac{1+ \frac{4}{625}(-5+s)^2}{ 1-\frac{1}{25}(-5+s)^2}}}ds \)


Edit: Corrected my Latex
Thanks for tacking a look!
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
Re: *tough* integrate

Hi Barioth, (Wave)

Your Latex syntax needs to be fixed a little bit it seems. I normally edit these things myself but I can't easily figure out what the integral is of so I don't want to guess. Can you try fixing the code or perhaps write it out in a different way so we can help you find the correct Latex code? :)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Hi everyone I have these 2 integrate that I can't solve, I have tried them with mathematica and wolfram, but they can't find an answer, maybe someone have an idea on how I could tackle these 2 bad boy!

The first one is

\(\displaystyle \int{ \sqrt{ \frac{1+( \frac{1}{10}+ \frac{s}{25})^2}{ \frac {s}{10}+ \frac{s^2}{50}}}}ds \)

The second one is

\(\displaystyle \int{ \sqrt{ \frac{1+ \frac{4}{625}(-5+s)^2}{ 1-\frac{1}{25}(-5+s)^2}}}ds \)


Edit: Corrected my Latex
Thanks for tacking a look!
They do not appear to succumb to any standard techniques. I suppose you could try differentiation under the integral sign, but it's not clear to me where to put the parameter. You'd have to play around with it some.

Could you provide some context for us? From where did these integrals appear?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Hi everyone I have these 2 integrate that I can't solve, I have tried them with mathematica and wolfram, but they can't find an answer, maybe someone have an idea on how I could tackle these 2 bad boy!

The first one is

\(\displaystyle \int{ \sqrt{ \frac{1+( \frac{1}{10}+ \frac{s}{25})^2}{ \frac {s}{10}+ \frac{s^2}{50}}}}ds \)

The second one is

\(\displaystyle \int{ \sqrt{ \frac{1+ \frac{4}{625}(-5+s)^2}{ 1-\frac{1}{25}(-5+s)^2}}}ds \)


Edit: Corrected my Latex
Thanks for tacking a look!
\(\displaystyle \displaystyle \begin{align*} \int{ \sqrt{\frac{1 + \left( \frac{1}{10} + \frac{s}{25} \right) ^2 }{\frac{s}{10} + \frac{s^2}{50}}} \,ds } &= \int{\left( \frac{\frac{1}{10} + \frac{s}{25}}{\frac{1}{10} + \frac{s}{25}} \right) \sqrt{ \frac{1 + \left( \frac{1}{10} + \frac{s}{25} \right) ^2 }{ \frac{s}{10} + \frac{s^2}{50} } }\,ds} \\ &= \int{\frac{\frac{1}{10} + \frac{s}{25}}{\sqrt{\frac{s}{10} + \frac{s^2}{50}}} \sqrt{ \frac{1 + \left( \frac{1}{10} + \frac{s}{25} \right)^2}{\left( \frac{1}{10} + \frac{s}{25} \right) ^2} } \,ds } \end{align*}\)

Now make the substitution \(\displaystyle \displaystyle u = \frac{s}{10} + \frac{s^2}{25} \implies du = \frac{1}{10} + \frac{s}{50}\,ds \) and note that

\(\displaystyle \displaystyle \begin{align*} \frac{s^2}{25} + \frac{s}{10} &= u \\ s^2 + \frac{5}{2}s &= 25u \\ s^2 + \frac{5}{2}s + \left( \frac{5}{4} \right) ^2 &= 25u + \left( \frac{5}{4} \right) ^2 \\ \left( s + \frac{5}{4} \right) ^2 &= \frac{400u + 25}{16} \\ s + \frac{5}{4} &= \pm \frac{\sqrt{400u + 25}}{4} \\ s &= \frac{-5 \pm 5\,\sqrt{16u + 1}}{4} \\ \frac{s}{25} &= \frac{-1 \pm \sqrt{16u + 1}}{20} \\ \frac{s}{25} + \frac{1}{10} &= \frac{1 \pm \sqrt{ 16u + 1}}{20} \end{align*}\)

Substituting into the integral gives

\(\displaystyle \displaystyle \begin{align*} \int{\frac{\frac{1}{10} + \frac{s}{25}}{\sqrt{\frac{s}{10} + \frac{s^2}{50}}} \sqrt{ \frac{1 + \left( \frac{1}{10} + \frac{s}{25} \right)^2}{\left( \frac{1}{10} + \frac{s}{25} \right) ^2} } \,ds } &= \int{ \frac{1}{\sqrt{u}} \sqrt{ \frac{ 1 + \left( \frac{1 \pm \sqrt{ 16u + 1}}{20} \right) ^2 }{ \left( \frac{1 \pm \sqrt{ 16u + 1}}{20} \right) ^2} } \,du} \end{align*}\)

I think I may have made it worse. Will need to keep thinking...
 

Barioth

Member
Jan 17, 2013
52
At first there was this:

\(\displaystyle \frac{dy}{ds}=\frac{1}{\sqrt{2}}* \sqrt{ \frac{1+( \frac{1}{10}+ \frac{s}{25})^2}{ \frac {s}{10}+ \frac{s^2}{50}}}\)

and this

\(\displaystyle \frac{dy}{ds}=\frac{1}{\sqrt{2}}*\sqrt{ \frac{1+ \frac{4}{625}(-5+s)^2}{ 1-\frac{1}{25}(-5+s)^2}}\)

I need a equation of s(y), So I tough I could just solve the integrate, but I'm starting to think that they may be no solution for it.

This come from

The time taken by a particule to do an infinitesimal path to the point (x+dx,y+dy) is given by

\(\displaystyle dt=\frac{\sqrt{dx^2+dy^2}}{\sqrt{2g(y_0-y)}}\)

Where x and y are the parametric parameter
with g= -1 and y0 = 0 I end up with

\(\displaystyle dt=\frac{\sqrt{dx^2+dy^2}}{\sqrt{2(y)}}\)

Maybe I should go with Diff Equation to solve this, but my skill in Diff Equations is near infinitesimal!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
The bit at the end looks like the brachistochrone problem. Are you trying to solve that problem? Perhaps you're trying to find $y$ as a function of the arc length $s$?
 

Barioth

Member
Jan 17, 2013
52
The bit at the end looks like the brachistochrone problem. Are you trying to solve that problem? Perhaps you're trying to find $y$ as a function of the arc length $s$?

Yeah I had to compare it with some other function, I asked my teacher about it, he told me I really had to use an aproximation (wich I really wanted to avoid.)

With an aproximation it get the job done, but it doesn't seem as beautiful to me..

Anyway thanks for taking some of your time guys!