Integral $int_{-pi/2}^{pi/2} frac{e^{|sin x|}cos x}{1+e^{tan x}}$












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This question was given in an eBook. I was checking out recently and I can't really find the answer to this question. Let me tell you that I am a bit of a novice on this topic of integration right now so I would request you to answer this accordingly.



$$int_{-pi/2}^{pi/2} frac{e^{|sin x|}cos x}{1+e^{tan x}}$$



So I first tried it using $$int_{b}^a f(x)dx=int_{b}^a f(a+b-x)dx$$ after splitting the integration in the limits $0tofrac{pi}{2}$ and $,-frac{pi}{2} to0$ because of the mod. Then I scratched it and tried integration by parts. But it seems everything just goes way outside the league of the question.



Well I hope you are able to solve it and help me. Cheers.






integration definite-integrals absolute-value







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    0












    $begingroup$


    This question was given in an eBook. I was checking out recently and I can't really find the answer to this question. Let me tell you that I am a bit of a novice on this topic of integration right now so I would request you to answer this accordingly.



    $$int_{-pi/2}^{pi/2} frac{e^{|sin x|}cos x}{1+e^{tan x}}$$



    So I first tried it using $$int_{b}^a f(x)dx=int_{b}^a f(a+b-x)dx$$ after splitting the integration in the limits $0tofrac{pi}{2}$ and $,-frac{pi}{2} to0$ because of the mod. Then I scratched it and tried integration by parts. But it seems everything just goes way outside the league of the question.



    Well I hope you are able to solve it and help me. Cheers.






    integration definite-integrals absolute-value







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      This question was given in an eBook. I was checking out recently and I can't really find the answer to this question. Let me tell you that I am a bit of a novice on this topic of integration right now so I would request you to answer this accordingly.



      $$int_{-pi/2}^{pi/2} frac{e^{|sin x|}cos x}{1+e^{tan x}}$$



      So I first tried it using $$int_{b}^a f(x)dx=int_{b}^a f(a+b-x)dx$$ after splitting the integration in the limits $0tofrac{pi}{2}$ and $,-frac{pi}{2} to0$ because of the mod. Then I scratched it and tried integration by parts. But it seems everything just goes way outside the league of the question.



      Well I hope you are able to solve it and help me. Cheers.






      integration definite-integrals absolute-value







      share|cite|improve this question











      $endgroup$




      This question was given in an eBook. I was checking out recently and I can't really find the answer to this question. Let me tell you that I am a bit of a novice on this topic of integration right now so I would request you to answer this accordingly.



      $$int_{-pi/2}^{pi/2} frac{e^{|sin x|}cos x}{1+e^{tan x}}$$



      So I first tried it using $$int_{b}^a f(x)dx=int_{b}^a f(a+b-x)dx$$ after splitting the integration in the limits $0tofrac{pi}{2}$ and $,-frac{pi}{2} to0$ because of the mod. Then I scratched it and tried integration by parts. But it seems everything just goes way outside the league of the question.



      Well I hope you are able to solve it and help me. Cheers.






      integration definite-integrals absolute-value




      integration definite-integrals absolute-value






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      share|cite|improve this question













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      edited Jan 9 at 18:15









      Zacky

      7,2551961




      7,2551961










      asked Aug 3 '18 at 12:19









      SriSri

      13910




      13910






















          2 Answers
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          $begingroup$

          with $x=-t$ $$I=int_{-pi/2}^{pi/2} frac{e^{|sin t|}cos t}{1+e^{-tan t}}mathrm dt$$ sum this with the initial integral and notice that $e^{|sin x|}cos x$ is an even function, therefore: $$I=2frac12int_{0}^{pi/2} e^{sin x} cos x mathrm dx$$ because $sin x$ is positive for $xin [0,frac{pi}{2}], $now just substitute $sin x= u$ to get: $$I=int_0^1 e^u du=e-1$$






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            2












            $begingroup$

            Hint:



            Here $a+b=0$



            $I=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin x|}cos x}{1+e^{tan x}}$



            $=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin(- x)|}cos(-x)}{1+e^{tan(-x)}}=int_{pi/2}^{-pi/2}dfrac{e^{tan x}e^{|sin x|}cos x}{1+e^{tan x}}$



            $$I+I=int_{pi/2}^{-pi/2}e^{|sin x|}cos x dx$$



            Now split from $-pi/2,0$ and $0,pi/2$






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              2 Answers
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              2 Answers
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              $begingroup$

              with $x=-t$ $$I=int_{-pi/2}^{pi/2} frac{e^{|sin t|}cos t}{1+e^{-tan t}}mathrm dt$$ sum this with the initial integral and notice that $e^{|sin x|}cos x$ is an even function, therefore: $$I=2frac12int_{0}^{pi/2} e^{sin x} cos x mathrm dx$$ because $sin x$ is positive for $xin [0,frac{pi}{2}], $now just substitute $sin x= u$ to get: $$I=int_0^1 e^u du=e-1$$






              share|cite|improve this answer









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                3












                $begingroup$

                with $x=-t$ $$I=int_{-pi/2}^{pi/2} frac{e^{|sin t|}cos t}{1+e^{-tan t}}mathrm dt$$ sum this with the initial integral and notice that $e^{|sin x|}cos x$ is an even function, therefore: $$I=2frac12int_{0}^{pi/2} e^{sin x} cos x mathrm dx$$ because $sin x$ is positive for $xin [0,frac{pi}{2}], $now just substitute $sin x= u$ to get: $$I=int_0^1 e^u du=e-1$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  with $x=-t$ $$I=int_{-pi/2}^{pi/2} frac{e^{|sin t|}cos t}{1+e^{-tan t}}mathrm dt$$ sum this with the initial integral and notice that $e^{|sin x|}cos x$ is an even function, therefore: $$I=2frac12int_{0}^{pi/2} e^{sin x} cos x mathrm dx$$ because $sin x$ is positive for $xin [0,frac{pi}{2}], $now just substitute $sin x= u$ to get: $$I=int_0^1 e^u du=e-1$$






                  share|cite|improve this answer









                  $endgroup$



                  with $x=-t$ $$I=int_{-pi/2}^{pi/2} frac{e^{|sin t|}cos t}{1+e^{-tan t}}mathrm dt$$ sum this with the initial integral and notice that $e^{|sin x|}cos x$ is an even function, therefore: $$I=2frac12int_{0}^{pi/2} e^{sin x} cos x mathrm dx$$ because $sin x$ is positive for $xin [0,frac{pi}{2}], $now just substitute $sin x= u$ to get: $$I=int_0^1 e^u du=e-1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 3 '18 at 12:28









                  ZackyZacky

                  7,2551961




                  7,2551961























                      2












                      $begingroup$

                      Hint:



                      Here $a+b=0$



                      $I=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin x|}cos x}{1+e^{tan x}}$



                      $=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin(- x)|}cos(-x)}{1+e^{tan(-x)}}=int_{pi/2}^{-pi/2}dfrac{e^{tan x}e^{|sin x|}cos x}{1+e^{tan x}}$



                      $$I+I=int_{pi/2}^{-pi/2}e^{|sin x|}cos x dx$$



                      Now split from $-pi/2,0$ and $0,pi/2$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Hint:



                        Here $a+b=0$



                        $I=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin x|}cos x}{1+e^{tan x}}$



                        $=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin(- x)|}cos(-x)}{1+e^{tan(-x)}}=int_{pi/2}^{-pi/2}dfrac{e^{tan x}e^{|sin x|}cos x}{1+e^{tan x}}$



                        $$I+I=int_{pi/2}^{-pi/2}e^{|sin x|}cos x dx$$



                        Now split from $-pi/2,0$ and $0,pi/2$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Hint:



                          Here $a+b=0$



                          $I=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin x|}cos x}{1+e^{tan x}}$



                          $=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin(- x)|}cos(-x)}{1+e^{tan(-x)}}=int_{pi/2}^{-pi/2}dfrac{e^{tan x}e^{|sin x|}cos x}{1+e^{tan x}}$



                          $$I+I=int_{pi/2}^{-pi/2}e^{|sin x|}cos x dx$$



                          Now split from $-pi/2,0$ and $0,pi/2$






                          share|cite|improve this answer









                          $endgroup$



                          Hint:



                          Here $a+b=0$



                          $I=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin x|}cos x}{1+e^{tan x}}$



                          $=displaystyleint_{pi/2}^{-pi/2}dfrac{e^{|sin(- x)|}cos(-x)}{1+e^{tan(-x)}}=int_{pi/2}^{-pi/2}dfrac{e^{tan x}e^{|sin x|}cos x}{1+e^{tan x}}$



                          $$I+I=int_{pi/2}^{-pi/2}e^{|sin x|}cos x dx$$



                          Now split from $-pi/2,0$ and $0,pi/2$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 3 '18 at 12:24









                          lab bhattacharjeelab bhattacharjee

                          226k15157275




                          226k15157275






























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