Integral $intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$












0












$begingroup$


I’m trying to solve this integral:
$$intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$$
As you can see it isn’t an easy integral to do in cartesian coordinates. However, given the fact that the numerator describes the upper half of the circumference $(x-b)^2 + y^2 = a^2$ this integral could be simpler in polar coordinates.



Yet I’m confused as to how can I use this idea. In polar coordinates $x = r cos (theta)$ so $text{d}x = cos(theta) text{d} r - rsin(theta) text{d} theta$. Do I have to substitute $x$ and $text{d}x$ in order to solve the integral in polar coordinates? Am I even right in thinking that polar coordinates should be used to solve this integral?






calculus definite-integrals







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Substitute $x=acos theta +b.$
    $endgroup$
    – John_Wick
    Nov 25 '18 at 6:51






  • 1




    $begingroup$
    Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:10










  • $begingroup$
    I’ll try this hint, thank you.
    $endgroup$
    – R.Mor
    Nov 25 '18 at 7:19










  • $begingroup$
    Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
    $endgroup$
    – David G. Stork
    Nov 25 '18 at 8:33












  • $begingroup$
    I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
    $endgroup$
    – R.Mor
    Nov 25 '18 at 15:04
















0












$begingroup$


I’m trying to solve this integral:
$$intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$$
As you can see it isn’t an easy integral to do in cartesian coordinates. However, given the fact that the numerator describes the upper half of the circumference $(x-b)^2 + y^2 = a^2$ this integral could be simpler in polar coordinates.



Yet I’m confused as to how can I use this idea. In polar coordinates $x = r cos (theta)$ so $text{d}x = cos(theta) text{d} r - rsin(theta) text{d} theta$. Do I have to substitute $x$ and $text{d}x$ in order to solve the integral in polar coordinates? Am I even right in thinking that polar coordinates should be used to solve this integral?






calculus definite-integrals







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Substitute $x=acos theta +b.$
    $endgroup$
    – John_Wick
    Nov 25 '18 at 6:51






  • 1




    $begingroup$
    Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:10










  • $begingroup$
    I’ll try this hint, thank you.
    $endgroup$
    – R.Mor
    Nov 25 '18 at 7:19










  • $begingroup$
    Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
    $endgroup$
    – David G. Stork
    Nov 25 '18 at 8:33












  • $begingroup$
    I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
    $endgroup$
    – R.Mor
    Nov 25 '18 at 15:04














0












0








0





$begingroup$


I’m trying to solve this integral:
$$intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$$
As you can see it isn’t an easy integral to do in cartesian coordinates. However, given the fact that the numerator describes the upper half of the circumference $(x-b)^2 + y^2 = a^2$ this integral could be simpler in polar coordinates.



Yet I’m confused as to how can I use this idea. In polar coordinates $x = r cos (theta)$ so $text{d}x = cos(theta) text{d} r - rsin(theta) text{d} theta$. Do I have to substitute $x$ and $text{d}x$ in order to solve the integral in polar coordinates? Am I even right in thinking that polar coordinates should be used to solve this integral?






calculus definite-integrals







share|cite|improve this question











$endgroup$




I’m trying to solve this integral:
$$intlimits_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x$$
As you can see it isn’t an easy integral to do in cartesian coordinates. However, given the fact that the numerator describes the upper half of the circumference $(x-b)^2 + y^2 = a^2$ this integral could be simpler in polar coordinates.



Yet I’m confused as to how can I use this idea. In polar coordinates $x = r cos (theta)$ so $text{d}x = cos(theta) text{d} r - rsin(theta) text{d} theta$. Do I have to substitute $x$ and $text{d}x$ in order to solve the integral in polar coordinates? Am I even right in thinking that polar coordinates should be used to solve this integral?






calculus definite-integrals




calculus definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 6:57









Arjang

5,58562363




5,58562363










asked Nov 25 '18 at 6:26









R.MorR.Mor

728




728








  • 3




    $begingroup$
    Substitute $x=acos theta +b.$
    $endgroup$
    – John_Wick
    Nov 25 '18 at 6:51






  • 1




    $begingroup$
    Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:10










  • $begingroup$
    I’ll try this hint, thank you.
    $endgroup$
    – R.Mor
    Nov 25 '18 at 7:19










  • $begingroup$
    Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
    $endgroup$
    – David G. Stork
    Nov 25 '18 at 8:33












  • $begingroup$
    I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
    $endgroup$
    – R.Mor
    Nov 25 '18 at 15:04














  • 3




    $begingroup$
    Substitute $x=acos theta +b.$
    $endgroup$
    – John_Wick
    Nov 25 '18 at 6:51






  • 1




    $begingroup$
    Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
    $endgroup$
    – Mason
    Nov 25 '18 at 7:10










  • $begingroup$
    I’ll try this hint, thank you.
    $endgroup$
    – R.Mor
    Nov 25 '18 at 7:19










  • $begingroup$
    Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
    $endgroup$
    – David G. Stork
    Nov 25 '18 at 8:33












  • $begingroup$
    I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
    $endgroup$
    – R.Mor
    Nov 25 '18 at 15:04








3




3




$begingroup$
Substitute $x=acos theta +b.$
$endgroup$
– John_Wick
Nov 25 '18 at 6:51




$begingroup$
Substitute $x=acos theta +b.$
$endgroup$
– John_Wick
Nov 25 '18 at 6:51




1




1




$begingroup$
Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
$endgroup$
– Mason
Nov 25 '18 at 7:10




$begingroup$
Note that this substitution is possible because $a cos theta +b$ has a range of $[b-a,b+a]$.
$endgroup$
– Mason
Nov 25 '18 at 7:10












$begingroup$
I’ll try this hint, thank you.
$endgroup$
– R.Mor
Nov 25 '18 at 7:19




$begingroup$
I’ll try this hint, thank you.
$endgroup$
– R.Mor
Nov 25 '18 at 7:19












$begingroup$
Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
$endgroup$
– David G. Stork
Nov 25 '18 at 8:33






$begingroup$
Assuming $a,b in mathbb{R}$ and $b>a$, Mathematica gives: $pi left(b-sqrt{b^2-a^2}right)$.
$endgroup$
– David G. Stork
Nov 25 '18 at 8:33














$begingroup$
I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
$endgroup$
– R.Mor
Nov 25 '18 at 15:04




$begingroup$
I plugged it in Mathematica as well (with the hypothesis you mentioned). However I was trying to solve it and it proved to be lengthy. Maybe in other coordinate system it was easier to do.
$endgroup$
– R.Mor
Nov 25 '18 at 15:04










1 Answer
1






active

oldest

votes


















2












$begingroup$

Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
    $endgroup$
    – R.Mor
    Nov 27 '18 at 15:28










  • $begingroup$
    You are welcome.
    $endgroup$
    – xpaul
    Nov 27 '18 at 19:39











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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
    $endgroup$
    – R.Mor
    Nov 27 '18 at 15:28










  • $begingroup$
    You are welcome.
    $endgroup$
    – xpaul
    Nov 27 '18 at 19:39
















2












$begingroup$

Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
    $endgroup$
    – R.Mor
    Nov 27 '18 at 15:28










  • $begingroup$
    You are welcome.
    $endgroup$
    – xpaul
    Nov 27 '18 at 19:39














2












2








2





$begingroup$

Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}






share|cite|improve this answer









$endgroup$



Suppose $b>a>0$. Let $x=b+asin t$ and $u=arctan(frac t2)$ then
begin{eqnarray*}
&&int_{b-a}^{b+a} frac{sqrt{a^2 - (x-b)^2}}{x} text{d} x\
&=&int_{-pi/2}^{pi/2}frac{a^2cos^2 t}{b+asin t}dt\
&=&2a^2int_{-1}^{1}frac{(u^2-1)^2}{(u^2+1)^2(bu^2+2au+b)}du\
&=&2a^2int_{-1}^{1}left[-frac{2u}{(u^2+1)^2}+frac{b}{u^2+1}+frac{a^2-b^2}{a^2(bu^2+2au+b)}right]du\
&=&2a^2left[frac{bpi}{2a^2}+frac{a^2-b^2}{a^2b}int_{-1}^{1}frac{1}{u^2+frac{2a}{b}u+1}duright]\
&=&2left[frac{bpi}{2}+frac{a^2-b^2}{b}int_{-1}^{1}frac{1}{(u+frac{a}{b})^2+frac{b^2-a^2}{b^2}}duright]\
&=&bpi+frac{2(a^2-b^2)}{b}frac{b}{sqrt{b^2-a^2}}arctanleft(frac{b(u+frac{a}{b})}{sqrt{b^2-a^2}}right)bigg|_{-1}^1\
&=&bpi-2sqrt{b^2-a^2}left(arctanleft(frac{b+a}{sqrt{b^2-a^2}}right)-arctanleft(frac{-b+a}{sqrt{b^2-a^2}}right)right)\
&=&bpi-2sqrt{b^2-a^2}cdotfrac{pi}{2}\
&=&pi(b-sqrt{b^2-a^2}).
end{eqnarray*}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 21:35









xpaulxpaul

22.6k24455




22.6k24455












  • $begingroup$
    I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
    $endgroup$
    – R.Mor
    Nov 27 '18 at 15:28










  • $begingroup$
    You are welcome.
    $endgroup$
    – xpaul
    Nov 27 '18 at 19:39


















  • $begingroup$
    I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
    $endgroup$
    – R.Mor
    Nov 27 '18 at 15:28










  • $begingroup$
    You are welcome.
    $endgroup$
    – xpaul
    Nov 27 '18 at 19:39
















$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28




$begingroup$
I arrived at the second integral (the one with the variable of integration $u$), then I got stuck. I didn't think of using partial fractions. Thank you for your answer
$endgroup$
– R.Mor
Nov 27 '18 at 15:28












$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39




$begingroup$
You are welcome.
$endgroup$
– xpaul
Nov 27 '18 at 19:39


















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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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