How to solve this definite integral $int_0^pi frac{cos^9(x)}{sin^3(x)+cos^3(x)}dx$












3














I'm having trouble evaluating the following integral:
$$
int^pi_0 frac{cos^9(x)}{sin^3(x)+cos^3(x)}dx
$$



I tried to convert it into an algebraic function by multiplying the numerator and denominator by $sec^{11}(x)$ and substituting $tan(x)=t$ as



$$
int^pi_0 frac{cos^9(x)cdot sec^{11}(x)}{(sin^3(x)+cos^3(x))cdot sec^{11}(x)}dx
$$



$$
int^pi_0 frac{sec^2(x)}{(tan^3(x)+1)cdot (tan^2(x)+1)^4}dx
$$



Substituting $tan(x)=t$,



$$
int^0_0 frac{dt}{(t^3+1)cdot (t^2+1)^4}
$$



But now both upper and lower limit become $0$ so apparently this is not the right approach, so how do I go about solving it?






integration definite-integrals







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  • 10




    Before you do anything, you need to decide what you want to do with the singularity at $x = frac{3pi}{4}$.
    – achille hui
    Jan 16 '16 at 19:20






  • 2




    One option is compute the Cauchy principal value instead. Another option is lift the integral to a contour integral in complex plane and deform the contour around the pole at $x = frac{3pi}{4}$. If this is the first time you deal with this sort of integral, you should follow the Cauchy principle value route. You essentially need to split the integral to 2 regions $[0,frac{3pi}{4}-epsilon], [frac{3pi}{4}+epsilon,pi]$ and take the limit $epsilon to 0^{+}$ at the end.
    – achille hui
    Jan 16 '16 at 19:32






  • 4




    Sure the integral is not on $[0,pi/2]$?
    – Did
    Jan 16 '16 at 20:19






  • 1




    "no harm in learning something new" Sure, and this is related how?
    – Did
    Jan 17 '16 at 10:34






  • 1




    When you do your substitutions, the functions in your substitutions must be continuous over the required interval. In this case, break it up into $(0, frac {pi}{2}), (frac {pi}{2}, pi)$ and you should be fine.
    – Doug M
    Dec 20 at 18:42


















3














I'm having trouble evaluating the following integral:
$$
int^pi_0 frac{cos^9(x)}{sin^3(x)+cos^3(x)}dx
$$



I tried to convert it into an algebraic function by multiplying the numerator and denominator by $sec^{11}(x)$ and substituting $tan(x)=t$ as



$$
int^pi_0 frac{cos^9(x)cdot sec^{11}(x)}{(sin^3(x)+cos^3(x))cdot sec^{11}(x)}dx
$$



$$
int^pi_0 frac{sec^2(x)}{(tan^3(x)+1)cdot (tan^2(x)+1)^4}dx
$$



Substituting $tan(x)=t$,



$$
int^0_0 frac{dt}{(t^3+1)cdot (t^2+1)^4}
$$



But now both upper and lower limit become $0$ so apparently this is not the right approach, so how do I go about solving it?






integration definite-integrals







share|cite|improve this question




















  • 10




    Before you do anything, you need to decide what you want to do with the singularity at $x = frac{3pi}{4}$.
    – achille hui
    Jan 16 '16 at 19:20






  • 2




    One option is compute the Cauchy principal value instead. Another option is lift the integral to a contour integral in complex plane and deform the contour around the pole at $x = frac{3pi}{4}$. If this is the first time you deal with this sort of integral, you should follow the Cauchy principle value route. You essentially need to split the integral to 2 regions $[0,frac{3pi}{4}-epsilon], [frac{3pi}{4}+epsilon,pi]$ and take the limit $epsilon to 0^{+}$ at the end.
    – achille hui
    Jan 16 '16 at 19:32






  • 4




    Sure the integral is not on $[0,pi/2]$?
    – Did
    Jan 16 '16 at 20:19






  • 1




    "no harm in learning something new" Sure, and this is related how?
    – Did
    Jan 17 '16 at 10:34






  • 1




    When you do your substitutions, the functions in your substitutions must be continuous over the required interval. In this case, break it up into $(0, frac {pi}{2}), (frac {pi}{2}, pi)$ and you should be fine.
    – Doug M
    Dec 20 at 18:42
















3












3








3


1





I'm having trouble evaluating the following integral:
$$
int^pi_0 frac{cos^9(x)}{sin^3(x)+cos^3(x)}dx
$$



I tried to convert it into an algebraic function by multiplying the numerator and denominator by $sec^{11}(x)$ and substituting $tan(x)=t$ as



$$
int^pi_0 frac{cos^9(x)cdot sec^{11}(x)}{(sin^3(x)+cos^3(x))cdot sec^{11}(x)}dx
$$



$$
int^pi_0 frac{sec^2(x)}{(tan^3(x)+1)cdot (tan^2(x)+1)^4}dx
$$



Substituting $tan(x)=t$,



$$
int^0_0 frac{dt}{(t^3+1)cdot (t^2+1)^4}
$$



But now both upper and lower limit become $0$ so apparently this is not the right approach, so how do I go about solving it?






integration definite-integrals







share|cite|improve this question















I'm having trouble evaluating the following integral:
$$
int^pi_0 frac{cos^9(x)}{sin^3(x)+cos^3(x)}dx
$$



I tried to convert it into an algebraic function by multiplying the numerator and denominator by $sec^{11}(x)$ and substituting $tan(x)=t$ as



$$
int^pi_0 frac{cos^9(x)cdot sec^{11}(x)}{(sin^3(x)+cos^3(x))cdot sec^{11}(x)}dx
$$



$$
int^pi_0 frac{sec^2(x)}{(tan^3(x)+1)cdot (tan^2(x)+1)^4}dx
$$



Substituting $tan(x)=t$,



$$
int^0_0 frac{dt}{(t^3+1)cdot (t^2+1)^4}
$$



But now both upper and lower limit become $0$ so apparently this is not the right approach, so how do I go about solving it?






integration definite-integrals




integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 27 '16 at 1:48









Martin Sleziak

44.7k7115270




44.7k7115270










asked Jan 16 '16 at 19:13









Apoorv

1436




1436








  • 10




    Before you do anything, you need to decide what you want to do with the singularity at $x = frac{3pi}{4}$.
    – achille hui
    Jan 16 '16 at 19:20






  • 2




    One option is compute the Cauchy principal value instead. Another option is lift the integral to a contour integral in complex plane and deform the contour around the pole at $x = frac{3pi}{4}$. If this is the first time you deal with this sort of integral, you should follow the Cauchy principle value route. You essentially need to split the integral to 2 regions $[0,frac{3pi}{4}-epsilon], [frac{3pi}{4}+epsilon,pi]$ and take the limit $epsilon to 0^{+}$ at the end.
    – achille hui
    Jan 16 '16 at 19:32






  • 4




    Sure the integral is not on $[0,pi/2]$?
    – Did
    Jan 16 '16 at 20:19






  • 1




    "no harm in learning something new" Sure, and this is related how?
    – Did
    Jan 17 '16 at 10:34






  • 1




    When you do your substitutions, the functions in your substitutions must be continuous over the required interval. In this case, break it up into $(0, frac {pi}{2}), (frac {pi}{2}, pi)$ and you should be fine.
    – Doug M
    Dec 20 at 18:42
















  • 10




    Before you do anything, you need to decide what you want to do with the singularity at $x = frac{3pi}{4}$.
    – achille hui
    Jan 16 '16 at 19:20






  • 2




    One option is compute the Cauchy principal value instead. Another option is lift the integral to a contour integral in complex plane and deform the contour around the pole at $x = frac{3pi}{4}$. If this is the first time you deal with this sort of integral, you should follow the Cauchy principle value route. You essentially need to split the integral to 2 regions $[0,frac{3pi}{4}-epsilon], [frac{3pi}{4}+epsilon,pi]$ and take the limit $epsilon to 0^{+}$ at the end.
    – achille hui
    Jan 16 '16 at 19:32






  • 4




    Sure the integral is not on $[0,pi/2]$?
    – Did
    Jan 16 '16 at 20:19






  • 1




    "no harm in learning something new" Sure, and this is related how?
    – Did
    Jan 17 '16 at 10:34






  • 1




    When you do your substitutions, the functions in your substitutions must be continuous over the required interval. In this case, break it up into $(0, frac {pi}{2}), (frac {pi}{2}, pi)$ and you should be fine.
    – Doug M
    Dec 20 at 18:42










10




10




Before you do anything, you need to decide what you want to do with the singularity at $x = frac{3pi}{4}$.
– achille hui
Jan 16 '16 at 19:20




Before you do anything, you need to decide what you want to do with the singularity at $x = frac{3pi}{4}$.
– achille hui
Jan 16 '16 at 19:20




2




2




One option is compute the Cauchy principal value instead. Another option is lift the integral to a contour integral in complex plane and deform the contour around the pole at $x = frac{3pi}{4}$. If this is the first time you deal with this sort of integral, you should follow the Cauchy principle value route. You essentially need to split the integral to 2 regions $[0,frac{3pi}{4}-epsilon], [frac{3pi}{4}+epsilon,pi]$ and take the limit $epsilon to 0^{+}$ at the end.
– achille hui
Jan 16 '16 at 19:32




One option is compute the Cauchy principal value instead. Another option is lift the integral to a contour integral in complex plane and deform the contour around the pole at $x = frac{3pi}{4}$. If this is the first time you deal with this sort of integral, you should follow the Cauchy principle value route. You essentially need to split the integral to 2 regions $[0,frac{3pi}{4}-epsilon], [frac{3pi}{4}+epsilon,pi]$ and take the limit $epsilon to 0^{+}$ at the end.
– achille hui
Jan 16 '16 at 19:32




4




4




Sure the integral is not on $[0,pi/2]$?
– Did
Jan 16 '16 at 20:19




Sure the integral is not on $[0,pi/2]$?
– Did
Jan 16 '16 at 20:19




1




1




"no harm in learning something new" Sure, and this is related how?
– Did
Jan 17 '16 at 10:34




"no harm in learning something new" Sure, and this is related how?
– Did
Jan 17 '16 at 10:34




1




1




When you do your substitutions, the functions in your substitutions must be continuous over the required interval. In this case, break it up into $(0, frac {pi}{2}), (frac {pi}{2}, pi)$ and you should be fine.
– Doug M
Dec 20 at 18:42






When you do your substitutions, the functions in your substitutions must be continuous over the required interval. In this case, break it up into $(0, frac {pi}{2}), (frac {pi}{2}, pi)$ and you should be fine.
– Doug M
Dec 20 at 18:42












1 Answer
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Assuming the usage of the Cauchy principle value, let's define
$$ I_1 := int_0^fracpi2 frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
\ I_2 := text{p. v.} int_fracpi2^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
$$



For $I_1$, by making a substitution and adding it to itself,
$$
I_1 = int_0^fracpi2 frac{sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
\ = frac12 int_0^fracpi2 frac{cos^9(x) +sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
\ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) - sin^3(x)cos^3(x) right) {rm d}x
$$



For $I_2$, by making some substitutions,
$$
I_2 = text{p. v.} int_0^fracpi2 frac{sin^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
\ = text{p. v.} int_0^fracpi2 frac{-cos^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
\ = frac12 text{p. v.} int_0^fracpi2 frac{sin^9(x)-cos^9(x)}{sin^3(x)-cos^3(x)}
\ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) + sin^3(x)cos^3(x) right) {rm d}x
$$



Therefore,



$$
I_1 + I_2 = text{p. v.} int_0^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
\ = int_0^fracpi2 sin^6(x) + cos^6(x) {rm d}x
\ = int_0^fracpi2 frac58 + frac38 cos(4x) {rm d}x
\ = frac{5pi}{16}
$$






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    Assuming the usage of the Cauchy principle value, let's define
    $$ I_1 := int_0^fracpi2 frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
    \ I_2 := text{p. v.} int_fracpi2^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
    $$



    For $I_1$, by making a substitution and adding it to itself,
    $$
    I_1 = int_0^fracpi2 frac{sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
    \ = frac12 int_0^fracpi2 frac{cos^9(x) +sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
    \ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) - sin^3(x)cos^3(x) right) {rm d}x
    $$



    For $I_2$, by making some substitutions,
    $$
    I_2 = text{p. v.} int_0^fracpi2 frac{sin^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
    \ = text{p. v.} int_0^fracpi2 frac{-cos^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
    \ = frac12 text{p. v.} int_0^fracpi2 frac{sin^9(x)-cos^9(x)}{sin^3(x)-cos^3(x)}
    \ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) + sin^3(x)cos^3(x) right) {rm d}x
    $$



    Therefore,



    $$
    I_1 + I_2 = text{p. v.} int_0^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
    \ = int_0^fracpi2 sin^6(x) + cos^6(x) {rm d}x
    \ = int_0^fracpi2 frac58 + frac38 cos(4x) {rm d}x
    \ = frac{5pi}{16}
    $$






    share|cite|improve this answer




























      1














      Assuming the usage of the Cauchy principle value, let's define
      $$ I_1 := int_0^fracpi2 frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
      \ I_2 := text{p. v.} int_fracpi2^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
      $$



      For $I_1$, by making a substitution and adding it to itself,
      $$
      I_1 = int_0^fracpi2 frac{sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
      \ = frac12 int_0^fracpi2 frac{cos^9(x) +sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
      \ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) - sin^3(x)cos^3(x) right) {rm d}x
      $$



      For $I_2$, by making some substitutions,
      $$
      I_2 = text{p. v.} int_0^fracpi2 frac{sin^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
      \ = text{p. v.} int_0^fracpi2 frac{-cos^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
      \ = frac12 text{p. v.} int_0^fracpi2 frac{sin^9(x)-cos^9(x)}{sin^3(x)-cos^3(x)}
      \ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) + sin^3(x)cos^3(x) right) {rm d}x
      $$



      Therefore,



      $$
      I_1 + I_2 = text{p. v.} int_0^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
      \ = int_0^fracpi2 sin^6(x) + cos^6(x) {rm d}x
      \ = int_0^fracpi2 frac58 + frac38 cos(4x) {rm d}x
      \ = frac{5pi}{16}
      $$






      share|cite|improve this answer


























        1












        1








        1






        Assuming the usage of the Cauchy principle value, let's define
        $$ I_1 := int_0^fracpi2 frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        \ I_2 := text{p. v.} int_fracpi2^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        $$



        For $I_1$, by making a substitution and adding it to itself,
        $$
        I_1 = int_0^fracpi2 frac{sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        \ = frac12 int_0^fracpi2 frac{cos^9(x) +sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        \ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) - sin^3(x)cos^3(x) right) {rm d}x
        $$



        For $I_2$, by making some substitutions,
        $$
        I_2 = text{p. v.} int_0^fracpi2 frac{sin^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
        \ = text{p. v.} int_0^fracpi2 frac{-cos^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
        \ = frac12 text{p. v.} int_0^fracpi2 frac{sin^9(x)-cos^9(x)}{sin^3(x)-cos^3(x)}
        \ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) + sin^3(x)cos^3(x) right) {rm d}x
        $$



        Therefore,



        $$
        I_1 + I_2 = text{p. v.} int_0^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        \ = int_0^fracpi2 sin^6(x) + cos^6(x) {rm d}x
        \ = int_0^fracpi2 frac58 + frac38 cos(4x) {rm d}x
        \ = frac{5pi}{16}
        $$






        share|cite|improve this answer














        Assuming the usage of the Cauchy principle value, let's define
        $$ I_1 := int_0^fracpi2 frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        \ I_2 := text{p. v.} int_fracpi2^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        $$



        For $I_1$, by making a substitution and adding it to itself,
        $$
        I_1 = int_0^fracpi2 frac{sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        \ = frac12 int_0^fracpi2 frac{cos^9(x) +sin^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        \ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) - sin^3(x)cos^3(x) right) {rm d}x
        $$



        For $I_2$, by making some substitutions,
        $$
        I_2 = text{p. v.} int_0^fracpi2 frac{sin^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
        \ = text{p. v.} int_0^fracpi2 frac{-cos^9(x)}{sin^3(x)-cos^3(x)} {rm d}x
        \ = frac12 text{p. v.} int_0^fracpi2 frac{sin^9(x)-cos^9(x)}{sin^3(x)-cos^3(x)}
        \ = frac12 int_0^fracpi2 left( sin^6(x) + cos^6(x) + sin^3(x)cos^3(x) right) {rm d}x
        $$



        Therefore,



        $$
        I_1 + I_2 = text{p. v.} int_0^pi frac{cos^9(x)}{cos^3(x)+sin^3(x)} {rm d}x
        \ = int_0^fracpi2 sin^6(x) + cos^6(x) {rm d}x
        \ = int_0^fracpi2 frac58 + frac38 cos(4x) {rm d}x
        \ = frac{5pi}{16}
        $$







        share|cite|improve this answer














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        edited Dec 26 at 15:32

























        answered Dec 20 at 18:25









        Mint

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            List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
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            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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