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Calculating this difficult integral (equation of a circle)

  1. Jul 22, 2009 #1
    I have tried numerous methods to calculate this integral and can't seem to figure it out.
    I have tried to use maple and mathematica but i am not very strong using these programs. I was wondering if someone would be able to help me solve this integral. This equation stemmed from the equation of a circle... and i got the following function:

    [tex]h = h0 + R - \sqrt{R+x}\sqrt{R-x}[/tex]

    i am trying to integrate the following:
    [tex]\int\frac{1}{h(x)^2}dx[/tex]

    and

    [tex]\int\frac{1}{h(x)^3}dx[/tex]

    Thanks for all your help

    MT
     
  2. jcsd
  3. Jul 22, 2009 #2
    Do you really need symbolic answers? For the first one, Maple says:
    [tex]
    -1/2\, \left( 2\,i{R}^{3}{\it h0}\,{\rm arctanh} \left( {\frac {{R}^{2
    }+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R \right)
    \sqrt {{R}^{2}-{x}^{2}}}} \right) +2\,i{R}^{3}{\it h0}\,{\rm arctanh}
    \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{
    \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -4\,{R}^{
    3}{\it h0}\,\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+
    2\,R}}} \right) +i{R}^{2}{{\it h0}}^{2}{\rm arctanh} \left( {\frac {{R
    }^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R
    \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) +i{R}^{2}{{\it h0}}^{2}{
    \rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{
    \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}
    \right) -2\,{R}^{2}\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{
    \it h0}+2\,R}}} \right) {{\it h0}}^{2}-2\,{R}^{2}x\sqrt {{\it h0}}
    \sqrt {{\it h0}+2\,R}+i{R}^{2}{\rm arctanh} \left( {\frac {-{R}^{2}+i
    \sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R \right)
    \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}+i{R}^{2}{\rm arctanh}
    \left( {\frac {{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{
    \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}-2
    \,{R}^{2}\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,
    R}}} \right) {x}^{2}-4\,Rx{{\it h0}}^{3/2}\sqrt {{\it h0}+2\,R}-2\,R
    \sqrt {R-x}\sqrt {R+x}\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}x-2\,x{{
    \it h0}}^{5/2}\sqrt {{\it h0}+2\,R}-2\,\sqrt {R-x}\sqrt {R+x}{{\it h0}
    }^{3/2}\sqrt {{\it h0}+2\,R}x \right) \left( {\it h0}+2\,R \right) ^{
    -3/2}{{\it h0}}^{-3/2} \left( {{\it h0}}^{2}+2\,{\it h0}\,R+{x}^{2}
    \right) ^{-1}
    [/tex]
     
  4. Jul 22, 2009 #3
    and for the second one, Maple says:
    [tex]
    1/4\, \left( -12\,i{R}^{4}{\it h0}\,{\rm arctanh} \left( {\frac {{R}^{
    2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R
    \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}-12\,i{R}^{4}{\it h0
    }\,{\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {
    {\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}
    \right) {x}^{2}-18\,i{R}^{3}{{\it h0}}^{2}{\rm arctanh} \left( {
    \frac {{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it
    h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}-18\,i{R}^{3}{{
    \it h0}}^{2}{\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,
    R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}
    }}} \right) {x}^{2}-6\,i{R}^{2}{{\it h0}}^{3}{\rm arctanh} \left( {
    \frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it
    h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}-6\,i{R}^{2}{{
    \it h0}}^{3}{\rm arctanh} \left( {\frac {{R}^{2}+i\sqrt {{\it h0}+2\,R
    }\sqrt {{\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}
    }} \right) {x}^{2}-3\,i{R}^{2}{\it h0}\,{\rm arctanh} \left( {\frac {{
    R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R
    \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{4}-3\,i{R}^{2}{\it h0}
    \,{\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{
    \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}
    \right) {x}^{4}+6\,\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{9/2}\sqrt {{\it
    h0}+2\,R}x+2\,\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{5/2}\sqrt {{\it h0}+2
    \,R}{x}^{3}+4\,{{\it h0}}^{11/2}\sqrt {{\it h0}+2\,R}x+24\,{R}^{5}
    \arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}}
    \right) {{\it h0}}^{2}+30\,{R}^{3}\arctan \left( {\frac {x}{\sqrt {{
    \it h0}}\sqrt {{\it h0}+2\,R}}} \right) {{\it h0}}^{4}+48\,{R}^{4}
    \arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}}
    \right) {{\it h0}}^{3}+6\,{R}^{3}\arctan \left( {\frac {x}{\sqrt {{
    \it h0}}\sqrt {{\it h0}+2\,R}}} \right) {x}^{4}+6\,{R}^{2}\arctan
    \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}} \right) {{
    \it h0}}^{5}+20\,{R}^{4}{{\it h0}}^{3/2}\sqrt {{\it h0}+2\,R}x+6\,{R}^
    {3}\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}{x}^{3}+46\,{R}^{3}{{\it h0}}^
    {5/2}\sqrt {{\it h0}+2\,R}x+42\,{R}^{2}{{\it h0}}^{7/2}\sqrt {{\it h0}
    +2\,R}x+6\,{R}^{2}{{\it h0}}^{3/2}\sqrt {{\it h0}+2\,R}{x}^{3}+20\,{{
    \it h0}}^{9/2}\sqrt {{\it h0}+2\,R}xR+36\,{R}^{3}\arctan \left( {
    \frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}} \right) {{\it h0}}^{
    2}{x}^{2}+6\,{R}^{2}\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{
    \it h0}+2\,R}}} \right) {\it h0}\,{x}^{4}+12\,{R}^{2}\arctan \left( {
    \frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}} \right) {{\it h0}}^{
    3}{x}^{2}+24\,{R}^{4}\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {
    {\it h0}+2\,R}}} \right) {\it h0}\,{x}^{2}-12\,i{R}^{5}{{\it h0}}^{2}{
    \rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{
    \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}
    \right) -12\,i{R}^{5}{{\it h0}}^{2}{\rm arctanh} \left( {\frac {{R}^{
    2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R
    \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -24\,i{R}^{4}{{\it h0}}^{3}
    {\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{
    \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}
    \right) -24\,i{R}^{4}{{\it h0}}^{3}{\rm arctanh} \left( {\frac {{R}^{
    2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R
    \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -15\,i{R}^{3}{{\it h0}}^{4}
    {\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{
    \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}
    \right) -15\,i{R}^{3}{{\it h0}}^{4}{\rm arctanh} \left( {\frac {{R}^{
    2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R
    \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -3\,i{R}^{3}{\rm arctanh}
    \left( {\frac {{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{
    \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{4}-3
    \,i{R}^{3}{\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}
    \sqrt {{\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}
    } \right) {x}^{4}-3\,i{R}^{2}{{\it h0}}^{5}{\rm arctanh} \left( {
    \frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it
    h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -3\,i{R}^{2}{{\it h0}}
    ^{5}{\rm arctanh} \left( {\frac {{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {
    {\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}
    \right) +20\,{R}^{3}\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{3/2}\sqrt {{
    \it h0}+2\,R}x+34\,{R}^{2}\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{5/2}\sqrt
    {{\it h0}+2\,R}x+24\,R\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{7/2}\sqrt {{
    \it h0}+2\,R}x+4\,R\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{3/2}\sqrt {{\it
    h0}+2\,R}{x}^{3}+6\,{R}^{2}\sqrt {R+x}\sqrt {R-x}\sqrt {{\it h0}}
    \sqrt {{\it h0}+2\,R}{x}^{3} \right) \left( -x+i\sqrt {{\it h0}+2\,R}
    \sqrt {{\it h0}} \right) ^{-2} \left( x+i\sqrt {{\it h0}+2\,R}\sqrt {{
    \it h0}} \right) ^{-2} \left( {\it h0}+2\,R \right) ^{-5/2}{{\it h0}}^
    {-5/2}
    [/tex]
    but this window doesn't show the whole thing in either case.
     
  5. Jul 22, 2009 #4
    unfortunately i do need a symbolic answer that i can use in matlab... every time i used maple i couldn't get an output i could copy and paste into matlab to use and i have no way to validate the solution from maple... thank you for your help!
     
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