Changing variable - Integration goes wrong.












2














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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  • I think this could help: math.stackexchange.com/questions/340180/…
    – Math-fun
    16 hours ago










  • The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
    – Kavi Rama Murthy
    16 hours ago










  • Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
    – Hans Lundmark
    16 hours ago
















2














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






















  • I think this could help: math.stackexchange.com/questions/340180/…
    – Math-fun
    16 hours ago










  • The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
    – Kavi Rama Murthy
    16 hours ago










  • Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
    – Hans Lundmark
    16 hours ago














2












2








2


1





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













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 17 hours ago









salvin

736




736












  • I think this could help: math.stackexchange.com/questions/340180/…
    – Math-fun
    16 hours ago










  • The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
    – Kavi Rama Murthy
    16 hours ago










  • Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
    – Hans Lundmark
    16 hours ago


















  • I think this could help: math.stackexchange.com/questions/340180/…
    – Math-fun
    16 hours ago










  • The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
    – Kavi Rama Murthy
    16 hours ago










  • Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
    – Hans Lundmark
    16 hours ago
















I think this could help: math.stackexchange.com/questions/340180/…
– Math-fun
16 hours ago




I think this could help: math.stackexchange.com/questions/340180/…
– Math-fun
16 hours ago












The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
– Kavi Rama Murthy
16 hours ago




The substitution $t=tan, x$ is not permitted. $tan$ is not nice enough for this (not even defined at $x=pi /2$).
– Kavi Rama Murthy
16 hours ago












Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
– Hans Lundmark
16 hours ago




Related (duplicates?): math.stackexchange.com/questions/829939/…, math.stackexchange.com/questions/2380669/…, and many others.
– Hans Lundmark
16 hours ago










2 Answers
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3














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





























    1














    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$$frac {pi}2intlimits_0^{infty}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}color{blue}{=frac {pi^2}{2ab}}$$






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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      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


























        3














        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
























          3












          3








          3






          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












          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 16 hours ago









          B. Goddard

          18.4k21340




          18.4k21340























              1














              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


























                1














                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
























                  1












                  1








                  1






                  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












                  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$$frac {pi}2intlimits_0^{infty}frac {mathrm dx}{a^2cos^2x+b^2sin^2x}color{blue}{=frac {pi^2}{2ab}}$$







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                  answered 5 hours ago









                  Frank W.

                  3,0811320




                  3,0811320






























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