Changing variable - Integration goes wrong.












1












$begingroup$


I was trying to do the integration
$$I=frac{pi}{2}int_0^pi frac{dx}{a^2cos^2x + b^2sin^2x}$$

If I divide throughout by $cos^2x$ and use substitution ($t=tan x$),

I obtain$$I=frac{pi}{2}int_0^0frac{dt}{a^2+(bt)^2} $$
This would be evaluated to be $0$.

But however its actual answer is $$frac{pi^2}{2ab}$$
which can be obtained by using properties of definite integrals to change the limit to $0$ to $frac{pi}{2}$ and then splitting it from $0$ to $frac{pi}{4}$ and $frac{pi}{4}$ to $frac{pi}{2}$ ,dividing throughout by $cos^2x$ and $sin^2x$ respectively and then substituting ($t=tan x$) and ($t=cot x$) respectively.

Can anyone please tell me what is wrong in the first approach.






calculus integration limits change-of-variable







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  • $begingroup$
    I think this could help: math.stackexchange.com/questions/340180/…
    $endgroup$
    – Math-fun
    Jan 4 at 12:34










  • $begingroup$
    The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 12:41










  • $begingroup$
    Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
    $endgroup$
    – Hans Lundmark
    Jan 4 at 12:42
















1












$begingroup$


I was trying to do the integration
$$I=frac{pi}{2}int_0^pi frac{dx}{a^2cos^2x + b^2sin^2x}$$

If I divide throughout by $cos^2x$ and use substitution ($t=tan x$),

I obtain$$I=frac{pi}{2}int_0^0frac{dt}{a^2+(bt)^2} $$
This would be evaluated to be $0$.

But however its actual answer is $$frac{pi^2}{2ab}$$
which can be obtained by using properties of definite integrals to change the limit to $0$ to $frac{pi}{2}$ and then splitting it from $0$ to $frac{pi}{4}$ and $frac{pi}{4}$ to $frac{pi}{2}$ ,dividing throughout by $cos^2x$ and $sin^2x$ respectively and then substituting ($t=tan x$) and ($t=cot x$) respectively.

Can anyone please tell me what is wrong in the first approach.






calculus integration limits change-of-variable







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think this could help: math.stackexchange.com/questions/340180/…
    $endgroup$
    – Math-fun
    Jan 4 at 12:34










  • $begingroup$
    The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 12:41










  • $begingroup$
    Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
    $endgroup$
    – Hans Lundmark
    Jan 4 at 12:42














1












1








1


1



$begingroup$


I was trying to do the integration
$$I=frac{pi}{2}int_0^pi frac{dx}{a^2cos^2x + b^2sin^2x}$$

If I divide throughout by $cos^2x$ and use substitution ($t=tan x$),

I obtain$$I=frac{pi}{2}int_0^0frac{dt}{a^2+(bt)^2} $$
This would be evaluated to be $0$.

But however its actual answer is $$frac{pi^2}{2ab}$$
which can be obtained by using properties of definite integrals to change the limit to $0$ to $frac{pi}{2}$ and then splitting it from $0$ to $frac{pi}{4}$ and $frac{pi}{4}$ to $frac{pi}{2}$ ,dividing throughout by $cos^2x$ and $sin^2x$ respectively and then substituting ($t=tan x$) and ($t=cot x$) respectively.

Can anyone please tell me what is wrong in the first approach.






calculus integration limits change-of-variable







share|cite|improve this question









$endgroup$




I was trying to do the integration
$$I=frac{pi}{2}int_0^pi frac{dx}{a^2cos^2x + b^2sin^2x}$$

If I divide throughout by $cos^2x$ and use substitution ($t=tan x$),

I obtain$$I=frac{pi}{2}int_0^0frac{dt}{a^2+(bt)^2} $$
This would be evaluated to be $0$.

But however its actual answer is $$frac{pi^2}{2ab}$$
which can be obtained by using properties of definite integrals to change the limit to $0$ to $frac{pi}{2}$ and then splitting it from $0$ to $frac{pi}{4}$ and $frac{pi}{4}$ to $frac{pi}{2}$ ,dividing throughout by $cos^2x$ and $sin^2x$ respectively and then substituting ($t=tan x$) and ($t=cot x$) respectively.

Can anyone please tell me what is wrong in the first approach.






calculus integration limits change-of-variable




calculus integration limits change-of-variable






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asked Jan 4 at 12:29









salvinsalvin

706




706












  • $begingroup$
    I think this could help: math.stackexchange.com/questions/340180/…
    $endgroup$
    – Math-fun
    Jan 4 at 12:34










  • $begingroup$
    The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 12:41










  • $begingroup$
    Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
    $endgroup$
    – Hans Lundmark
    Jan 4 at 12:42


















  • $begingroup$
    I think this could help: math.stackexchange.com/questions/340180/…
    $endgroup$
    – Math-fun
    Jan 4 at 12:34










  • $begingroup$
    The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
    $endgroup$
    – Kavi Rama Murthy
    Jan 4 at 12:41










  • $begingroup$
    Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
    $endgroup$
    – Hans Lundmark
    Jan 4 at 12:42
















$begingroup$
I think this could help: math.stackexchange.com/questions/340180/…
$endgroup$
– Math-fun
Jan 4 at 12:34




$begingroup$
I think this could help: math.stackexchange.com/questions/340180/…
$endgroup$
– Math-fun
Jan 4 at 12:34












$begingroup$
The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
$endgroup$
– Kavi Rama Murthy
Jan 4 at 12:41




$begingroup$
The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
$endgroup$
– Kavi Rama Murthy
Jan 4 at 12:41












$begingroup$
Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
$endgroup$
– Hans Lundmark
Jan 4 at 12:42




$begingroup$
Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
$endgroup$
– Hans Lundmark
Jan 4 at 12:42










2 Answers
2






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oldest

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3












$begingroup$

The reason why your substitution doesn't work is due to the singularity at $frac {pi}2$. Indeed, $tanleft(frac {pi}2right)$ isn't defined. However, you can remedy this by splitting up the integral along the singularity$$begin{align*}intlimits_0^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x} & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_{pi/2}^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}\ & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_0^{pi/2}frac {mathrm dx}{a^2sin^2x+b^2cos^2x}\ & =intlimits_0^{infty}frac {mathrm dx}{a^2+b^2x^2}+intlimits_0^{infty}frac {mathrm dx}{b^2+a^2x^2}\ & =frac 1{ab}left[arctanleft(frac {bx}aright)+arctanleft(frac {ax}bright)right],Biggrrvert_0^{infty}end{align*}$$So now, we get that$$frac {pi}2intlimits_0^{infty}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}color{blue}{=frac {pi^2}{2ab}}$$






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    The main problem is that is that $tan t$ has a singularity at $pi/2$. In $u$-substitution, the function $u$ is required to be differentiable on the interval and it's not even continuous here.



    See the Wiki article on $u$-substitution:



    https://en.wikipedia.org/wiki/Integration_by_substitution






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      3












      $begingroup$

      The reason why your substitution doesn't work is due to the singularity at $frac {pi}2$. Indeed, $tanleft(frac {pi}2right)$ isn't defined. However, you can remedy this by splitting up the integral along the singularity$$begin{align*}intlimits_0^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x} & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_{pi/2}^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}\ & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_0^{pi/2}frac {mathrm dx}{a^2sin^2x+b^2cos^2x}\ & =intlimits_0^{infty}frac {mathrm dx}{a^2+b^2x^2}+intlimits_0^{infty}frac {mathrm dx}{b^2+a^2x^2}\ & =frac 1{ab}left[arctanleft(frac {bx}aright)+arctanleft(frac {ax}bright)right],Biggrrvert_0^{infty}end{align*}$$So now, we get that$$frac {pi}2intlimits_0^{infty}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}color{blue}{=frac {pi^2}{2ab}}$$






      share|cite|improve this answer









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        3












        $begingroup$

        The reason why your substitution doesn't work is due to the singularity at $frac {pi}2$. Indeed, $tanleft(frac {pi}2right)$ isn't defined. However, you can remedy this by splitting up the integral along the singularity$$begin{align*}intlimits_0^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x} & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_{pi/2}^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}\ & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_0^{pi/2}frac {mathrm dx}{a^2sin^2x+b^2cos^2x}\ & =intlimits_0^{infty}frac {mathrm dx}{a^2+b^2x^2}+intlimits_0^{infty}frac {mathrm dx}{b^2+a^2x^2}\ & =frac 1{ab}left[arctanleft(frac {bx}aright)+arctanleft(frac {ax}bright)right],Biggrrvert_0^{infty}end{align*}$$So now, we get that$$frac {pi}2intlimits_0^{infty}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}color{blue}{=frac {pi^2}{2ab}}$$






        share|cite|improve this answer









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          3












          3








          3





          $begingroup$

          The reason why your substitution doesn't work is due to the singularity at $frac {pi}2$. Indeed, $tanleft(frac {pi}2right)$ isn't defined. However, you can remedy this by splitting up the integral along the singularity$$begin{align*}intlimits_0^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x} & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_{pi/2}^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}\ & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_0^{pi/2}frac {mathrm dx}{a^2sin^2x+b^2cos^2x}\ & =intlimits_0^{infty}frac {mathrm dx}{a^2+b^2x^2}+intlimits_0^{infty}frac {mathrm dx}{b^2+a^2x^2}\ & =frac 1{ab}left[arctanleft(frac {bx}aright)+arctanleft(frac {ax}bright)right],Biggrrvert_0^{infty}end{align*}$$So now, we get that$$frac {pi}2intlimits_0^{infty}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}color{blue}{=frac {pi^2}{2ab}}$$






          share|cite|improve this answer









          $endgroup$



          The reason why your substitution doesn't work is due to the singularity at $frac {pi}2$. Indeed, $tanleft(frac {pi}2right)$ isn't defined. However, you can remedy this by splitting up the integral along the singularity$$begin{align*}intlimits_0^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x} & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_{pi/2}^{pi}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}\ & =intlimits_0^{pi/2}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}+intlimits_0^{pi/2}frac {mathrm dx}{a^2sin^2x+b^2cos^2x}\ & =intlimits_0^{infty}frac {mathrm dx}{a^2+b^2x^2}+intlimits_0^{infty}frac {mathrm dx}{b^2+a^2x^2}\ & =frac 1{ab}left[arctanleft(frac {bx}aright)+arctanleft(frac {ax}bright)right],Biggrrvert_0^{infty}end{align*}$$So now, we get that$$frac {pi}2intlimits_0^{infty}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}color{blue}{=frac {pi^2}{2ab}}$$







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Jan 5 at 0:23









          Frank W.Frank W.

          3,8001321




          3,8001321























              6












              $begingroup$

              The main problem is that is that $tan t$ has a singularity at $pi/2$. In $u$-substitution, the function $u$ is required to be differentiable on the interval and it's not even continuous here.



              See the Wiki article on $u$-substitution:



              https://en.wikipedia.org/wiki/Integration_by_substitution






              share|cite|improve this answer









              $endgroup$


















                6












                $begingroup$

                The main problem is that is that $tan t$ has a singularity at $pi/2$. In $u$-substitution, the function $u$ is required to be differentiable on the interval and it's not even continuous here.



                See the Wiki article on $u$-substitution:



                https://en.wikipedia.org/wiki/Integration_by_substitution






                share|cite|improve this answer









                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  The main problem is that is that $tan t$ has a singularity at $pi/2$. In $u$-substitution, the function $u$ is required to be differentiable on the interval and it's not even continuous here.



                  See the Wiki article on $u$-substitution:



                  https://en.wikipedia.org/wiki/Integration_by_substitution






                  share|cite|improve this answer









                  $endgroup$



                  The main problem is that is that $tan t$ has a singularity at $pi/2$. In $u$-substitution, the function $u$ is required to be differentiable on the interval and it's not even continuous here.



                  See the Wiki article on $u$-substitution:



                  https://en.wikipedia.org/wiki/Integration_by_substitution







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 12:40









                  B. GoddardB. Goddard

                  19.6k21442




                  19.6k21442






























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