What's the answer to $int frac{cos^2x sin x}{sin x - cos x} dx$?












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I tried solving the integral $$int frac{cos^2x sin x}{sin x - cos x}, dx$$ the following ways:




  1. Expressing each function in the form of $tan left(frac{x}{2}right)$, $cos left(frac{x}{2}right),$ and $,sin left(frac{x}{2}right),$ independently, but that didn't go well for me.


  2. Multiplying and dividing by $cos^2x$ or $sin^2x$.


  3. Expressing $cos^2x$ as $1-sin^2x$ and splitting the integral, and I was stuck with $int left(frac{sin^3x}{sin x - cos x}right), dx$ which I rewrote as $int frac{csc^2x}{csc^4x (1-cot x) } dx,,$ and tried a whole range of substitutions only to fail.


  4. I tried to substitute $frac{1}{ sin x - cos x}$, $frac{sin x}{ sin x - cos x}$, $frac{cos x sin x}{ sin x - cos x}$ and $frac{cos^2x sin x}{sin x - cos x},$ independently, none of which seemed to work out.


  5. I expressed the denominator as $sinleft(frac{pi}{4}-xright)$ and tried multiplying and dividing by $sinleft(frac{pi}{4}+xright)$, and carried out some substitutions. Then, I repeated the same with $cosleft(frac{pi}{4}+xright)$. Neither of them worked.







real-analysis integration indefinite-integrals substitution trigonometric-integrals







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    4












    $begingroup$


    I tried solving the integral $$int frac{cos^2x sin x}{sin x - cos x}, dx$$ the following ways:




    1. Expressing each function in the form of $tan left(frac{x}{2}right)$, $cos left(frac{x}{2}right),$ and $,sin left(frac{x}{2}right),$ independently, but that didn't go well for me.


    2. Multiplying and dividing by $cos^2x$ or $sin^2x$.


    3. Expressing $cos^2x$ as $1-sin^2x$ and splitting the integral, and I was stuck with $int left(frac{sin^3x}{sin x - cos x}right), dx$ which I rewrote as $int frac{csc^2x}{csc^4x (1-cot x) } dx,,$ and tried a whole range of substitutions only to fail.


    4. I tried to substitute $frac{1}{ sin x - cos x}$, $frac{sin x}{ sin x - cos x}$, $frac{cos x sin x}{ sin x - cos x}$ and $frac{cos^2x sin x}{sin x - cos x},$ independently, none of which seemed to work out.


    5. I expressed the denominator as $sinleft(frac{pi}{4}-xright)$ and tried multiplying and dividing by $sinleft(frac{pi}{4}+xright)$, and carried out some substitutions. Then, I repeated the same with $cosleft(frac{pi}{4}+xright)$. Neither of them worked.







    real-analysis integration indefinite-integrals substitution trigonometric-integrals







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I tried solving the integral $$int frac{cos^2x sin x}{sin x - cos x}, dx$$ the following ways:




      1. Expressing each function in the form of $tan left(frac{x}{2}right)$, $cos left(frac{x}{2}right),$ and $,sin left(frac{x}{2}right),$ independently, but that didn't go well for me.


      2. Multiplying and dividing by $cos^2x$ or $sin^2x$.


      3. Expressing $cos^2x$ as $1-sin^2x$ and splitting the integral, and I was stuck with $int left(frac{sin^3x}{sin x - cos x}right), dx$ which I rewrote as $int frac{csc^2x}{csc^4x (1-cot x) } dx,,$ and tried a whole range of substitutions only to fail.


      4. I tried to substitute $frac{1}{ sin x - cos x}$, $frac{sin x}{ sin x - cos x}$, $frac{cos x sin x}{ sin x - cos x}$ and $frac{cos^2x sin x}{sin x - cos x},$ independently, none of which seemed to work out.


      5. I expressed the denominator as $sinleft(frac{pi}{4}-xright)$ and tried multiplying and dividing by $sinleft(frac{pi}{4}+xright)$, and carried out some substitutions. Then, I repeated the same with $cosleft(frac{pi}{4}+xright)$. Neither of them worked.







      real-analysis integration indefinite-integrals substitution trigonometric-integrals







      share|cite|improve this question











      $endgroup$




      I tried solving the integral $$int frac{cos^2x sin x}{sin x - cos x}, dx$$ the following ways:




      1. Expressing each function in the form of $tan left(frac{x}{2}right)$, $cos left(frac{x}{2}right),$ and $,sin left(frac{x}{2}right),$ independently, but that didn't go well for me.


      2. Multiplying and dividing by $cos^2x$ or $sin^2x$.


      3. Expressing $cos^2x$ as $1-sin^2x$ and splitting the integral, and I was stuck with $int left(frac{sin^3x}{sin x - cos x}right), dx$ which I rewrote as $int frac{csc^2x}{csc^4x (1-cot x) } dx,,$ and tried a whole range of substitutions only to fail.


      4. I tried to substitute $frac{1}{ sin x - cos x}$, $frac{sin x}{ sin x - cos x}$, $frac{cos x sin x}{ sin x - cos x}$ and $frac{cos^2x sin x}{sin x - cos x},$ independently, none of which seemed to work out.


      5. I expressed the denominator as $sinleft(frac{pi}{4}-xright)$ and tried multiplying and dividing by $sinleft(frac{pi}{4}+xright)$, and carried out some substitutions. Then, I repeated the same with $cosleft(frac{pi}{4}+xright)$. Neither of them worked.







      real-analysis integration indefinite-integrals substitution trigonometric-integrals




      real-analysis integration indefinite-integrals substitution trigonometric-integrals






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 21 at 18:47









      Michael Rozenberg

      107k1894199




      107k1894199










      asked Jan 21 at 17:22









      Sashank SriramSashank Sriram

      444




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          6 Answers
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          $$intfrac{cos^2xsin{x}}{sin{x}-cos{x}}dx=$$
          $$=intleft(frac{cos^2xsin{x}}{sin{x}-cos{x}}+frac{1}{2}sin{x}(sin{x}+cos{x})right)dx-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
          $$=frac{1}{2}intfrac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
          $$=frac{1}{2}intleft(frac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}right)dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
          $$=frac{1}{4}intfrac{sin{x}+cos{x}}{sin{x}-cos{x}}dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
          $$=frac{1}{4}ln|sin{x}-cos{x}|-frac{1}{4}intleft(2sin^2{x}-1+sin2xright)dx.$$
          Can you end it now?






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            3












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            Hint:



            Let $dfracpi4-x=y$



            $sin x-cos x=sqrt2sin y$



            $sin x=dfrac{cos y-sin y}{sqrt2}$



            $2cos^2x=1+cos2x=1+2sin ycos y$



            $$dfrac{(cos y-sin y)(1+2sin ycos y)}{sin y}=2cos^2y-2sin ycos y+cot y-1=cos2y-sin2y+cot y$$






            share|cite|improve this answer









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              3












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              Let $displaystyle I =frac{1}{2}intfrac{2cos^2 xcdot sin x}{sin x-cos x}dx=frac{1}{2}intfrac{(1+cos 2x)cdot sin x}{sin x-cos x}dx$



              So $displaystyle I =frac{1}{4}int frac{2sin x}{sin x-cos x}dx+frac{1}{2}intfrac{cos 2xcdot sin x}{sin x-cos x}dx$



              Now writting



              $2sin x=(sin x+cos x)+(sin x-cos x)$



              and $cos (2x)=cos^2 x-sin^2 x.$






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                $2cos^2 x=1+cos2x $.
                $endgroup$
                – Thomas Shelby
                Jan 21 at 17:37



















              2












              $begingroup$

              A standard substitution for integrals of rational functions of trigonometric ones is $t=tan frac{x}{2}$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                No, doesn't work
                $endgroup$
                – Sashank Sriram
                Jan 21 at 17:46










              • $begingroup$
                It does though.
                $endgroup$
                – lightxbulb
                Jan 21 at 17:54










              • $begingroup$
                Then learn how to integrate rational functions: sosmath.com/calculus/integration/rational/rational.html @Sashank Sriram
                $endgroup$
                – lightxbulb
                Jan 21 at 18:00





















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              try it with $$sin(x)=frac{2t}{1+t^2}$$
              $$cos(x)=frac{1-t^2}{1+t^2}$$
              $$dx=frac{2dt}{1+t^2}$$
              You will get this integral $$int frac{4 t
              left(t^2-1right)^2}{left(t^2+1
              right)^3 left(t^2+2 t-1right)}dt$$
              and this is $$int left(1/2,{frac {-t+1}{{t}^{2}+1}}+{frac {-4,t+4}{ left( {t}^{2}+1
              right) ^{3}}}+1/2,{frac {t+1}{{t}^{2}+2,t-1}}+{frac {2,t-4}{
              left( {t}^{2}+1 right) ^{2}}}right)
              dt$$






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              $endgroup$













              • $begingroup$
                Going by that method got me to a messy fraction, that lead me nowhere
                $endgroup$
                – Sashank Sriram
                Jan 21 at 17:48



















              1












              $begingroup$

              For integrands that are a rational function of sine and cosine, as a guide to what trigonometric substitution one may try the Bioche Rules can be used.



              Here notice the differential form
              $$w(x) = f(sin x, cos x) , dx = frac{cos^2 x sin x}{sin x - cos x} , dx$$
              is invariant under the substitution $x mapsto pi + x$, that is, $w(pi + x) = w(x)$. This suggests a substitution of $t = tan x$ can be used. As $dt = sec^2 x , dx$ we rewrite the integrand as the product between a rational function consisting of $tan x$ terms and a $sec^2 x$ terms. Doing so we have
              begin{align}
              I &= int frac{cos^2 x sin x}{sin x - cos x} , dx\
              &= int frac{cos^2 x sin x}{sin x - cos x} cdot frac{sec^2 x}{sec^2 x} , dx\
              &= int frac{tan x}{(tan x - 1)(1 + tan^2 x)^2} cdot sec^2 x , dx.
              end{align}
              Now let $t = tan x$. Doing so yields
              begin{align}
              I &= int frac{t}{(t - 1)(1 + t^2)^2} , dt\
              &= int left [frac{1}{4(t - 1)} - frac{t + 1}{4(t^2 + 1)} + frac{1 - t}{2(1 + t^2)^2} right ] , dt\
              &= frac{1}{4} ln |t - 1| - frac{1}{8} ln |1 + t^2| + frac{t}{4(1 + t^2)} + C\
              &= frac{1}{4}ln |tan x - 1| + frac{1}{4} ln |cos x| + frac{1}{4} (1 + tan x) cos^2 x + C.
              end{align}
              or after playing around with a few trignometric identities
              $$int frac{cos^2 x sin x}{sin x - cos x} , dx = frac{1}{4} ln |sin x - cos x| + frac{1}{8} (cos 2x + sin 2x) + C.$$






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                6 Answers
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                5












                $begingroup$

                $$intfrac{cos^2xsin{x}}{sin{x}-cos{x}}dx=$$
                $$=intleft(frac{cos^2xsin{x}}{sin{x}-cos{x}}+frac{1}{2}sin{x}(sin{x}+cos{x})right)dx-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
                $$=frac{1}{2}intfrac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
                $$=frac{1}{2}intleft(frac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}right)dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
                $$=frac{1}{4}intfrac{sin{x}+cos{x}}{sin{x}-cos{x}}dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
                $$=frac{1}{4}ln|sin{x}-cos{x}|-frac{1}{4}intleft(2sin^2{x}-1+sin2xright)dx.$$
                Can you end it now?






                share|cite|improve this answer









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                  5












                  $begingroup$

                  $$intfrac{cos^2xsin{x}}{sin{x}-cos{x}}dx=$$
                  $$=intleft(frac{cos^2xsin{x}}{sin{x}-cos{x}}+frac{1}{2}sin{x}(sin{x}+cos{x})right)dx-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
                  $$=frac{1}{2}intfrac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
                  $$=frac{1}{2}intleft(frac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}right)dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
                  $$=frac{1}{4}intfrac{sin{x}+cos{x}}{sin{x}-cos{x}}dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
                  $$=frac{1}{4}ln|sin{x}-cos{x}|-frac{1}{4}intleft(2sin^2{x}-1+sin2xright)dx.$$
                  Can you end it now?






                  share|cite|improve this answer









                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    $$intfrac{cos^2xsin{x}}{sin{x}-cos{x}}dx=$$
                    $$=intleft(frac{cos^2xsin{x}}{sin{x}-cos{x}}+frac{1}{2}sin{x}(sin{x}+cos{x})right)dx-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
                    $$=frac{1}{2}intfrac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
                    $$=frac{1}{2}intleft(frac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}right)dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
                    $$=frac{1}{4}intfrac{sin{x}+cos{x}}{sin{x}-cos{x}}dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
                    $$=frac{1}{4}ln|sin{x}-cos{x}|-frac{1}{4}intleft(2sin^2{x}-1+sin2xright)dx.$$
                    Can you end it now?






                    share|cite|improve this answer









                    $endgroup$



                    $$intfrac{cos^2xsin{x}}{sin{x}-cos{x}}dx=$$
                    $$=intleft(frac{cos^2xsin{x}}{sin{x}-cos{x}}+frac{1}{2}sin{x}(sin{x}+cos{x})right)dx-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
                    $$=frac{1}{2}intfrac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}intsin{x}(sin{x}+cos{x})dx=$$
                    $$=frac{1}{2}intleft(frac{sin{x}}{sin{x}-cos{x}}-frac{1}{2}right)dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
                    $$=frac{1}{4}intfrac{sin{x}+cos{x}}{sin{x}-cos{x}}dx-frac{1}{2}intleft(sin{x}(sin{x}+cos{x})-frac{1}{2}right)dx=$$
                    $$=frac{1}{4}ln|sin{x}-cos{x}|-frac{1}{4}intleft(2sin^2{x}-1+sin2xright)dx.$$
                    Can you end it now?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 21 at 17:56









                    Michael RozenbergMichael Rozenberg

                    107k1894199




                    107k1894199























                        3












                        $begingroup$

                        Hint:



                        Let $dfracpi4-x=y$



                        $sin x-cos x=sqrt2sin y$



                        $sin x=dfrac{cos y-sin y}{sqrt2}$



                        $2cos^2x=1+cos2x=1+2sin ycos y$



                        $$dfrac{(cos y-sin y)(1+2sin ycos y)}{sin y}=2cos^2y-2sin ycos y+cot y-1=cos2y-sin2y+cot y$$






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          Hint:



                          Let $dfracpi4-x=y$



                          $sin x-cos x=sqrt2sin y$



                          $sin x=dfrac{cos y-sin y}{sqrt2}$



                          $2cos^2x=1+cos2x=1+2sin ycos y$



                          $$dfrac{(cos y-sin y)(1+2sin ycos y)}{sin y}=2cos^2y-2sin ycos y+cot y-1=cos2y-sin2y+cot y$$






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            Hint:



                            Let $dfracpi4-x=y$



                            $sin x-cos x=sqrt2sin y$



                            $sin x=dfrac{cos y-sin y}{sqrt2}$



                            $2cos^2x=1+cos2x=1+2sin ycos y$



                            $$dfrac{(cos y-sin y)(1+2sin ycos y)}{sin y}=2cos^2y-2sin ycos y+cot y-1=cos2y-sin2y+cot y$$






                            share|cite|improve this answer









                            $endgroup$



                            Hint:



                            Let $dfracpi4-x=y$



                            $sin x-cos x=sqrt2sin y$



                            $sin x=dfrac{cos y-sin y}{sqrt2}$



                            $2cos^2x=1+cos2x=1+2sin ycos y$



                            $$dfrac{(cos y-sin y)(1+2sin ycos y)}{sin y}=2cos^2y-2sin ycos y+cot y-1=cos2y-sin2y+cot y$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 21 at 18:30









                            lab bhattacharjeelab bhattacharjee

                            226k15157275




                            226k15157275























                                3












                                $begingroup$

                                Let $displaystyle I =frac{1}{2}intfrac{2cos^2 xcdot sin x}{sin x-cos x}dx=frac{1}{2}intfrac{(1+cos 2x)cdot sin x}{sin x-cos x}dx$



                                So $displaystyle I =frac{1}{4}int frac{2sin x}{sin x-cos x}dx+frac{1}{2}intfrac{cos 2xcdot sin x}{sin x-cos x}dx$



                                Now writting



                                $2sin x=(sin x+cos x)+(sin x-cos x)$



                                and $cos (2x)=cos^2 x-sin^2 x.$






                                share|cite|improve this answer











                                $endgroup$









                                • 1




                                  $begingroup$
                                  $2cos^2 x=1+cos2x $.
                                  $endgroup$
                                  – Thomas Shelby
                                  Jan 21 at 17:37
















                                3












                                $begingroup$

                                Let $displaystyle I =frac{1}{2}intfrac{2cos^2 xcdot sin x}{sin x-cos x}dx=frac{1}{2}intfrac{(1+cos 2x)cdot sin x}{sin x-cos x}dx$



                                So $displaystyle I =frac{1}{4}int frac{2sin x}{sin x-cos x}dx+frac{1}{2}intfrac{cos 2xcdot sin x}{sin x-cos x}dx$



                                Now writting



                                $2sin x=(sin x+cos x)+(sin x-cos x)$



                                and $cos (2x)=cos^2 x-sin^2 x.$






                                share|cite|improve this answer











                                $endgroup$









                                • 1




                                  $begingroup$
                                  $2cos^2 x=1+cos2x $.
                                  $endgroup$
                                  – Thomas Shelby
                                  Jan 21 at 17:37














                                3












                                3








                                3





                                $begingroup$

                                Let $displaystyle I =frac{1}{2}intfrac{2cos^2 xcdot sin x}{sin x-cos x}dx=frac{1}{2}intfrac{(1+cos 2x)cdot sin x}{sin x-cos x}dx$



                                So $displaystyle I =frac{1}{4}int frac{2sin x}{sin x-cos x}dx+frac{1}{2}intfrac{cos 2xcdot sin x}{sin x-cos x}dx$



                                Now writting



                                $2sin x=(sin x+cos x)+(sin x-cos x)$



                                and $cos (2x)=cos^2 x-sin^2 x.$






                                share|cite|improve this answer











                                $endgroup$



                                Let $displaystyle I =frac{1}{2}intfrac{2cos^2 xcdot sin x}{sin x-cos x}dx=frac{1}{2}intfrac{(1+cos 2x)cdot sin x}{sin x-cos x}dx$



                                So $displaystyle I =frac{1}{4}int frac{2sin x}{sin x-cos x}dx+frac{1}{2}intfrac{cos 2xcdot sin x}{sin x-cos x}dx$



                                Now writting



                                $2sin x=(sin x+cos x)+(sin x-cos x)$



                                and $cos (2x)=cos^2 x-sin^2 x.$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jan 21 at 21:18

























                                answered Jan 21 at 17:30









                                DXTDXT

                                5,9992732




                                5,9992732








                                • 1




                                  $begingroup$
                                  $2cos^2 x=1+cos2x $.
                                  $endgroup$
                                  – Thomas Shelby
                                  Jan 21 at 17:37














                                • 1




                                  $begingroup$
                                  $2cos^2 x=1+cos2x $.
                                  $endgroup$
                                  – Thomas Shelby
                                  Jan 21 at 17:37








                                1




                                1




                                $begingroup$
                                $2cos^2 x=1+cos2x $.
                                $endgroup$
                                – Thomas Shelby
                                Jan 21 at 17:37




                                $begingroup$
                                $2cos^2 x=1+cos2x $.
                                $endgroup$
                                – Thomas Shelby
                                Jan 21 at 17:37











                                2












                                $begingroup$

                                A standard substitution for integrals of rational functions of trigonometric ones is $t=tan frac{x}{2}$.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  No, doesn't work
                                  $endgroup$
                                  – Sashank Sriram
                                  Jan 21 at 17:46










                                • $begingroup$
                                  It does though.
                                  $endgroup$
                                  – lightxbulb
                                  Jan 21 at 17:54










                                • $begingroup$
                                  Then learn how to integrate rational functions: sosmath.com/calculus/integration/rational/rational.html @Sashank Sriram
                                  $endgroup$
                                  – lightxbulb
                                  Jan 21 at 18:00


















                                2












                                $begingroup$

                                A standard substitution for integrals of rational functions of trigonometric ones is $t=tan frac{x}{2}$.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  No, doesn't work
                                  $endgroup$
                                  – Sashank Sriram
                                  Jan 21 at 17:46










                                • $begingroup$
                                  It does though.
                                  $endgroup$
                                  – lightxbulb
                                  Jan 21 at 17:54










                                • $begingroup$
                                  Then learn how to integrate rational functions: sosmath.com/calculus/integration/rational/rational.html @Sashank Sriram
                                  $endgroup$
                                  – lightxbulb
                                  Jan 21 at 18:00
















                                2












                                2








                                2





                                $begingroup$

                                A standard substitution for integrals of rational functions of trigonometric ones is $t=tan frac{x}{2}$.






                                share|cite|improve this answer









                                $endgroup$



                                A standard substitution for integrals of rational functions of trigonometric ones is $t=tan frac{x}{2}$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 21 at 17:32









                                lightxbulblightxbulb

                                1,055311




                                1,055311












                                • $begingroup$
                                  No, doesn't work
                                  $endgroup$
                                  – Sashank Sriram
                                  Jan 21 at 17:46










                                • $begingroup$
                                  It does though.
                                  $endgroup$
                                  – lightxbulb
                                  Jan 21 at 17:54










                                • $begingroup$
                                  Then learn how to integrate rational functions: sosmath.com/calculus/integration/rational/rational.html @Sashank Sriram
                                  $endgroup$
                                  – lightxbulb
                                  Jan 21 at 18:00




















                                • $begingroup$
                                  No, doesn't work
                                  $endgroup$
                                  – Sashank Sriram
                                  Jan 21 at 17:46










                                • $begingroup$
                                  It does though.
                                  $endgroup$
                                  – lightxbulb
                                  Jan 21 at 17:54










                                • $begingroup$
                                  Then learn how to integrate rational functions: sosmath.com/calculus/integration/rational/rational.html @Sashank Sriram
                                  $endgroup$
                                  – lightxbulb
                                  Jan 21 at 18:00


















                                $begingroup$
                                No, doesn't work
                                $endgroup$
                                – Sashank Sriram
                                Jan 21 at 17:46




                                $begingroup$
                                No, doesn't work
                                $endgroup$
                                – Sashank Sriram
                                Jan 21 at 17:46












                                $begingroup$
                                It does though.
                                $endgroup$
                                – lightxbulb
                                Jan 21 at 17:54




                                $begingroup$
                                It does though.
                                $endgroup$
                                – lightxbulb
                                Jan 21 at 17:54












                                $begingroup$
                                Then learn how to integrate rational functions: sosmath.com/calculus/integration/rational/rational.html @Sashank Sriram
                                $endgroup$
                                – lightxbulb
                                Jan 21 at 18:00






                                $begingroup$
                                Then learn how to integrate rational functions: sosmath.com/calculus/integration/rational/rational.html @Sashank Sriram
                                $endgroup$
                                – lightxbulb
                                Jan 21 at 18:00













                                2












                                $begingroup$

                                try it with $$sin(x)=frac{2t}{1+t^2}$$
                                $$cos(x)=frac{1-t^2}{1+t^2}$$
                                $$dx=frac{2dt}{1+t^2}$$
                                You will get this integral $$int frac{4 t
                                left(t^2-1right)^2}{left(t^2+1
                                right)^3 left(t^2+2 t-1right)}dt$$
                                and this is $$int left(1/2,{frac {-t+1}{{t}^{2}+1}}+{frac {-4,t+4}{ left( {t}^{2}+1
                                right) ^{3}}}+1/2,{frac {t+1}{{t}^{2}+2,t-1}}+{frac {2,t-4}{
                                left( {t}^{2}+1 right) ^{2}}}right)
                                dt$$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Going by that method got me to a messy fraction, that lead me nowhere
                                  $endgroup$
                                  – Sashank Sriram
                                  Jan 21 at 17:48
















                                2












                                $begingroup$

                                try it with $$sin(x)=frac{2t}{1+t^2}$$
                                $$cos(x)=frac{1-t^2}{1+t^2}$$
                                $$dx=frac{2dt}{1+t^2}$$
                                You will get this integral $$int frac{4 t
                                left(t^2-1right)^2}{left(t^2+1
                                right)^3 left(t^2+2 t-1right)}dt$$
                                and this is $$int left(1/2,{frac {-t+1}{{t}^{2}+1}}+{frac {-4,t+4}{ left( {t}^{2}+1
                                right) ^{3}}}+1/2,{frac {t+1}{{t}^{2}+2,t-1}}+{frac {2,t-4}{
                                left( {t}^{2}+1 right) ^{2}}}right)
                                dt$$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Going by that method got me to a messy fraction, that lead me nowhere
                                  $endgroup$
                                  – Sashank Sriram
                                  Jan 21 at 17:48














                                2












                                2








                                2





                                $begingroup$

                                try it with $$sin(x)=frac{2t}{1+t^2}$$
                                $$cos(x)=frac{1-t^2}{1+t^2}$$
                                $$dx=frac{2dt}{1+t^2}$$
                                You will get this integral $$int frac{4 t
                                left(t^2-1right)^2}{left(t^2+1
                                right)^3 left(t^2+2 t-1right)}dt$$
                                and this is $$int left(1/2,{frac {-t+1}{{t}^{2}+1}}+{frac {-4,t+4}{ left( {t}^{2}+1
                                right) ^{3}}}+1/2,{frac {t+1}{{t}^{2}+2,t-1}}+{frac {2,t-4}{
                                left( {t}^{2}+1 right) ^{2}}}right)
                                dt$$






                                share|cite|improve this answer











                                $endgroup$



                                try it with $$sin(x)=frac{2t}{1+t^2}$$
                                $$cos(x)=frac{1-t^2}{1+t^2}$$
                                $$dx=frac{2dt}{1+t^2}$$
                                You will get this integral $$int frac{4 t
                                left(t^2-1right)^2}{left(t^2+1
                                right)^3 left(t^2+2 t-1right)}dt$$
                                and this is $$int left(1/2,{frac {-t+1}{{t}^{2}+1}}+{frac {-4,t+4}{ left( {t}^{2}+1
                                right) ^{3}}}+1/2,{frac {t+1}{{t}^{2}+2,t-1}}+{frac {2,t-4}{
                                left( {t}^{2}+1 right) ^{2}}}right)
                                dt$$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Jan 21 at 18:07

























                                answered Jan 21 at 17:31









                                Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                                77k42866




                                77k42866












                                • $begingroup$
                                  Going by that method got me to a messy fraction, that lead me nowhere
                                  $endgroup$
                                  – Sashank Sriram
                                  Jan 21 at 17:48


















                                • $begingroup$
                                  Going by that method got me to a messy fraction, that lead me nowhere
                                  $endgroup$
                                  – Sashank Sriram
                                  Jan 21 at 17:48
















                                $begingroup$
                                Going by that method got me to a messy fraction, that lead me nowhere
                                $endgroup$
                                – Sashank Sriram
                                Jan 21 at 17:48




                                $begingroup$
                                Going by that method got me to a messy fraction, that lead me nowhere
                                $endgroup$
                                – Sashank Sriram
                                Jan 21 at 17:48











                                1












                                $begingroup$

                                For integrands that are a rational function of sine and cosine, as a guide to what trigonometric substitution one may try the Bioche Rules can be used.



                                Here notice the differential form
                                $$w(x) = f(sin x, cos x) , dx = frac{cos^2 x sin x}{sin x - cos x} , dx$$
                                is invariant under the substitution $x mapsto pi + x$, that is, $w(pi + x) = w(x)$. This suggests a substitution of $t = tan x$ can be used. As $dt = sec^2 x , dx$ we rewrite the integrand as the product between a rational function consisting of $tan x$ terms and a $sec^2 x$ terms. Doing so we have
                                begin{align}
                                I &= int frac{cos^2 x sin x}{sin x - cos x} , dx\
                                &= int frac{cos^2 x sin x}{sin x - cos x} cdot frac{sec^2 x}{sec^2 x} , dx\
                                &= int frac{tan x}{(tan x - 1)(1 + tan^2 x)^2} cdot sec^2 x , dx.
                                end{align}
                                Now let $t = tan x$. Doing so yields
                                begin{align}
                                I &= int frac{t}{(t - 1)(1 + t^2)^2} , dt\
                                &= int left [frac{1}{4(t - 1)} - frac{t + 1}{4(t^2 + 1)} + frac{1 - t}{2(1 + t^2)^2} right ] , dt\
                                &= frac{1}{4} ln |t - 1| - frac{1}{8} ln |1 + t^2| + frac{t}{4(1 + t^2)} + C\
                                &= frac{1}{4}ln |tan x - 1| + frac{1}{4} ln |cos x| + frac{1}{4} (1 + tan x) cos^2 x + C.
                                end{align}
                                or after playing around with a few trignometric identities
                                $$int frac{cos^2 x sin x}{sin x - cos x} , dx = frac{1}{4} ln |sin x - cos x| + frac{1}{8} (cos 2x + sin 2x) + C.$$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  For integrands that are a rational function of sine and cosine, as a guide to what trigonometric substitution one may try the Bioche Rules can be used.



                                  Here notice the differential form
                                  $$w(x) = f(sin x, cos x) , dx = frac{cos^2 x sin x}{sin x - cos x} , dx$$
                                  is invariant under the substitution $x mapsto pi + x$, that is, $w(pi + x) = w(x)$. This suggests a substitution of $t = tan x$ can be used. As $dt = sec^2 x , dx$ we rewrite the integrand as the product between a rational function consisting of $tan x$ terms and a $sec^2 x$ terms. Doing so we have
                                  begin{align}
                                  I &= int frac{cos^2 x sin x}{sin x - cos x} , dx\
                                  &= int frac{cos^2 x sin x}{sin x - cos x} cdot frac{sec^2 x}{sec^2 x} , dx\
                                  &= int frac{tan x}{(tan x - 1)(1 + tan^2 x)^2} cdot sec^2 x , dx.
                                  end{align}
                                  Now let $t = tan x$. Doing so yields
                                  begin{align}
                                  I &= int frac{t}{(t - 1)(1 + t^2)^2} , dt\
                                  &= int left [frac{1}{4(t - 1)} - frac{t + 1}{4(t^2 + 1)} + frac{1 - t}{2(1 + t^2)^2} right ] , dt\
                                  &= frac{1}{4} ln |t - 1| - frac{1}{8} ln |1 + t^2| + frac{t}{4(1 + t^2)} + C\
                                  &= frac{1}{4}ln |tan x - 1| + frac{1}{4} ln |cos x| + frac{1}{4} (1 + tan x) cos^2 x + C.
                                  end{align}
                                  or after playing around with a few trignometric identities
                                  $$int frac{cos^2 x sin x}{sin x - cos x} , dx = frac{1}{4} ln |sin x - cos x| + frac{1}{8} (cos 2x + sin 2x) + C.$$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    For integrands that are a rational function of sine and cosine, as a guide to what trigonometric substitution one may try the Bioche Rules can be used.



                                    Here notice the differential form
                                    $$w(x) = f(sin x, cos x) , dx = frac{cos^2 x sin x}{sin x - cos x} , dx$$
                                    is invariant under the substitution $x mapsto pi + x$, that is, $w(pi + x) = w(x)$. This suggests a substitution of $t = tan x$ can be used. As $dt = sec^2 x , dx$ we rewrite the integrand as the product between a rational function consisting of $tan x$ terms and a $sec^2 x$ terms. Doing so we have
                                    begin{align}
                                    I &= int frac{cos^2 x sin x}{sin x - cos x} , dx\
                                    &= int frac{cos^2 x sin x}{sin x - cos x} cdot frac{sec^2 x}{sec^2 x} , dx\
                                    &= int frac{tan x}{(tan x - 1)(1 + tan^2 x)^2} cdot sec^2 x , dx.
                                    end{align}
                                    Now let $t = tan x$. Doing so yields
                                    begin{align}
                                    I &= int frac{t}{(t - 1)(1 + t^2)^2} , dt\
                                    &= int left [frac{1}{4(t - 1)} - frac{t + 1}{4(t^2 + 1)} + frac{1 - t}{2(1 + t^2)^2} right ] , dt\
                                    &= frac{1}{4} ln |t - 1| - frac{1}{8} ln |1 + t^2| + frac{t}{4(1 + t^2)} + C\
                                    &= frac{1}{4}ln |tan x - 1| + frac{1}{4} ln |cos x| + frac{1}{4} (1 + tan x) cos^2 x + C.
                                    end{align}
                                    or after playing around with a few trignometric identities
                                    $$int frac{cos^2 x sin x}{sin x - cos x} , dx = frac{1}{4} ln |sin x - cos x| + frac{1}{8} (cos 2x + sin 2x) + C.$$






                                    share|cite|improve this answer









                                    $endgroup$



                                    For integrands that are a rational function of sine and cosine, as a guide to what trigonometric substitution one may try the Bioche Rules can be used.



                                    Here notice the differential form
                                    $$w(x) = f(sin x, cos x) , dx = frac{cos^2 x sin x}{sin x - cos x} , dx$$
                                    is invariant under the substitution $x mapsto pi + x$, that is, $w(pi + x) = w(x)$. This suggests a substitution of $t = tan x$ can be used. As $dt = sec^2 x , dx$ we rewrite the integrand as the product between a rational function consisting of $tan x$ terms and a $sec^2 x$ terms. Doing so we have
                                    begin{align}
                                    I &= int frac{cos^2 x sin x}{sin x - cos x} , dx\
                                    &= int frac{cos^2 x sin x}{sin x - cos x} cdot frac{sec^2 x}{sec^2 x} , dx\
                                    &= int frac{tan x}{(tan x - 1)(1 + tan^2 x)^2} cdot sec^2 x , dx.
                                    end{align}
                                    Now let $t = tan x$. Doing so yields
                                    begin{align}
                                    I &= int frac{t}{(t - 1)(1 + t^2)^2} , dt\
                                    &= int left [frac{1}{4(t - 1)} - frac{t + 1}{4(t^2 + 1)} + frac{1 - t}{2(1 + t^2)^2} right ] , dt\
                                    &= frac{1}{4} ln |t - 1| - frac{1}{8} ln |1 + t^2| + frac{t}{4(1 + t^2)} + C\
                                    &= frac{1}{4}ln |tan x - 1| + frac{1}{4} ln |cos x| + frac{1}{4} (1 + tan x) cos^2 x + C.
                                    end{align}
                                    or after playing around with a few trignometric identities
                                    $$int frac{cos^2 x sin x}{sin x - cos x} , dx = frac{1}{4} ln |sin x - cos x| + frac{1}{8} (cos 2x + sin 2x) + C.$$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 22 at 2:48









                                    omegadotomegadot

                                    6,4172829




                                    6,4172829






























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