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












0














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




















  • 3




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






  • 1




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










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










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












  • 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.
    – R.Mor
    Nov 25 at 15:04
















0














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




















  • 3




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






  • 1




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










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










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












  • 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.
    – R.Mor
    Nov 25 at 15:04














0












0








0







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















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 at 6:57









Arjang

5,56362363




5,56362363










asked Nov 25 at 6:26









R.Mor

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  • 3




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






  • 1




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










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










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












  • 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.
    – R.Mor
    Nov 25 at 15:04














  • 3




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






  • 1




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










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










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












  • 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.
    – R.Mor
    Nov 25 at 15:04








3




3




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




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




1




1




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




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












I’ll try this hint, thank you.
– R.Mor
Nov 25 at 7:19




I’ll try this hint, thank you.
– R.Mor
Nov 25 at 7:19












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






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














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.
– R.Mor
Nov 25 at 15:04




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.
– R.Mor
Nov 25 at 15:04










1 Answer
1






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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





















  • 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
    – R.Mor
    Nov 27 at 15:28










  • You are welcome.
    – xpaul
    Nov 27 at 19:39











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1 Answer
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1 Answer
1






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active

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active

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2














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





















  • 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
    – R.Mor
    Nov 27 at 15:28










  • You are welcome.
    – xpaul
    Nov 27 at 19:39
















2














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





















  • 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
    – R.Mor
    Nov 27 at 15:28










  • You are welcome.
    – xpaul
    Nov 27 at 19:39














2












2








2






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












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 at 21:35









xpaul

22.4k14455




22.4k14455












  • 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
    – R.Mor
    Nov 27 at 15:28










  • You are welcome.
    – xpaul
    Nov 27 at 19:39


















  • 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
    – R.Mor
    Nov 27 at 15:28










  • You are welcome.
    – xpaul
    Nov 27 at 19:39
















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
– R.Mor
Nov 27 at 15:28




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
– R.Mor
Nov 27 at 15:28












You are welcome.
– xpaul
Nov 27 at 19:39




You are welcome.
– xpaul
Nov 27 at 19:39


















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