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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012501%2fintegral-int-limits-b-aba-frac-sqrta2-x-b2x-textd-x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3012501%2fintegral-int-limits-b-aba-frac-sqrta2-x-b2x-textd-x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
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
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more