Can these integrals be solved by u-substitution using trig identities?












0














$int frac{operatorname{sin} x}{2sin xcos x+5}dx$



$int frac{operatorname{sin}xcos x}{operatorname{sin}x+cos x}dx$



$int frac{dx}{operatorname{sin} x+cos x}$



I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:



$int frac{operatorname{sin}2x(operatorname{sin}x-cos x)}{cos 2x}dx$ but didn't know how to go from there.






calculus integration trigonometry indefinite-integrals trigonometric-integrals







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  • Single question please
    – lab bhattacharjee
    Nov 24 at 16:01










  • Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
    – DavidG
    Dec 9 at 10:46
















0














$int frac{operatorname{sin} x}{2sin xcos x+5}dx$



$int frac{operatorname{sin}xcos x}{operatorname{sin}x+cos x}dx$



$int frac{dx}{operatorname{sin} x+cos x}$



I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:



$int frac{operatorname{sin}2x(operatorname{sin}x-cos x)}{cos 2x}dx$ but didn't know how to go from there.






calculus integration trigonometry indefinite-integrals trigonometric-integrals







share|cite|improve this question
























  • Single question please
    – lab bhattacharjee
    Nov 24 at 16:01










  • Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
    – DavidG
    Dec 9 at 10:46














0












0








0


0





$int frac{operatorname{sin} x}{2sin xcos x+5}dx$



$int frac{operatorname{sin}xcos x}{operatorname{sin}x+cos x}dx$



$int frac{dx}{operatorname{sin} x+cos x}$



I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:



$int frac{operatorname{sin}2x(operatorname{sin}x-cos x)}{cos 2x}dx$ but didn't know how to go from there.






calculus integration trigonometry indefinite-integrals trigonometric-integrals







share|cite|improve this question















$int frac{operatorname{sin} x}{2sin xcos x+5}dx$



$int frac{operatorname{sin}xcos x}{operatorname{sin}x+cos x}dx$



$int frac{dx}{operatorname{sin} x+cos x}$



I'd like to know if there's a way to solve some of these integrals by manipulating it and then doing u substitution, I tried but I'm not very good at trig identities. I got the second one like this:



$int frac{operatorname{sin}2x(operatorname{sin}x-cos x)}{cos 2x}dx$ but didn't know how to go from there.






calculus integration trigonometry indefinite-integrals trigonometric-integrals




calculus integration trigonometry indefinite-integrals trigonometric-integrals






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edited Nov 25 at 17:32









Martin Sleziak

44.6k7115270




44.6k7115270










asked Nov 24 at 15:57









Andres Oropeza

355




355












  • Single question please
    – lab bhattacharjee
    Nov 24 at 16:01










  • Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
    – DavidG
    Dec 9 at 10:46


















  • Single question please
    – lab bhattacharjee
    Nov 24 at 16:01










  • Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
    – DavidG
    Dec 9 at 10:46
















Single question please
– lab bhattacharjee
Nov 24 at 16:01




Single question please
– lab bhattacharjee
Nov 24 at 16:01












Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
– DavidG
Dec 9 at 10:46




Integrals of these types are quite often easily solved using the Weierstrass Substitution (en.wikipedia.org/wiki/Tangent_half-angle_substitution).
– DavidG
Dec 9 at 10:46










3 Answers
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1














Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$






share|cite|improve this answer





























    2














    Hint:



    $$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$



    Set $x+pi/4=y$



    $cos2y=1-2sin^2y$






    share|cite|improve this answer





























      2














      For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$



      For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$



      For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$






      share|cite|improve this answer























      • Could you expand a Little bit?
        – Andres Oropeza
        Nov 24 at 19:21










      • @AndresOropeza, Please find the updated answer
        – lab bhattacharjee
        Nov 25 at 16:34











      Your Answer





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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

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      active

      oldest

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      1














      Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$






      share|cite|improve this answer


























        1














        Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$






        share|cite|improve this answer
























          1












          1








          1






          Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$






          share|cite|improve this answer












          Assuming $operatorname{sen}$ is synonymous with $sin$, the third integral is $$inttfrac{1}{sqrt{2}}csc (x+tfrac{pi}{4})dx=-tfrac{1}{sqrt{2}}ln |csc (x+tfrac{pi}{4})+cot (x+tfrac{pi}{4})|+C=-tfrac{1}{sqrt{2}}ln |cot (tfrac{x}{2}+tfrac{pi}{8})|+C,$$while the first integral is very messy (although integrating it between certain limits might be neater). Just to expand on lab bhattacharjee's treatment of the second integral,$$intfrac{1}{sqrt{8}}frac{sin (2y-tfrac{pi}{2})}{sin y}dy=-intfrac{1}{sqrt{8}}frac{cos 2y}{sin y}dy=intfrac{2sin y-csc y}{sqrt{8}}dy\=frac{-2cos y+ln|csc y+cot y|}{sqrt{8}}+C=frac{-2cos (x+tfrac{pi}{4})+ln|cot (tfrac{x}{2}+tfrac{pi}{8})|}{sqrt{8}}+C.$$







          share|cite|improve this answer












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          share|cite|improve this answer










          answered Nov 24 at 16:21









          J.G.

          21.7k21934




          21.7k21934























              2














              Hint:



              $$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$



              Set $x+pi/4=y$



              $cos2y=1-2sin^2y$






              share|cite|improve this answer


























                2














                Hint:



                $$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$



                Set $x+pi/4=y$



                $cos2y=1-2sin^2y$






                share|cite|improve this answer
























                  2












                  2








                  2






                  Hint:



                  $$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$



                  Set $x+pi/4=y$



                  $cos2y=1-2sin^2y$






                  share|cite|improve this answer












                  Hint:



                  $$dfrac{sin xcos x}{sin x+cos x}=dfrac{sin2x}{2sqrt2sin(x+pi/4)}$$



                  Set $x+pi/4=y$



                  $cos2y=1-2sin^2y$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 16:05









                  lab bhattacharjee

                  222k15155273




                  222k15155273























                      2














                      For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$



                      For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$



                      For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$






                      share|cite|improve this answer























                      • Could you expand a Little bit?
                        – Andres Oropeza
                        Nov 24 at 19:21










                      • @AndresOropeza, Please find the updated answer
                        – lab bhattacharjee
                        Nov 25 at 16:34
















                      2














                      For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$



                      For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$



                      For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$






                      share|cite|improve this answer























                      • Could you expand a Little bit?
                        – Andres Oropeza
                        Nov 24 at 19:21










                      • @AndresOropeza, Please find the updated answer
                        – lab bhattacharjee
                        Nov 25 at 16:34














                      2












                      2








                      2






                      For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$



                      For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$



                      For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$






                      share|cite|improve this answer














                      For the first one, write numerator as $$dfrac{2sin x}{2sin xcos x+5}=dfrac{sin x-cos x}{2(2sin xcos x+5)}+dfrac{sin x+cos x}{2(2sin xcos x+5)}$$



                      For the first part, write $$int(sin x-cos x)dx=u, u^2=1+2sin xcos xiffsin xcos x=?$$



                      For the second, $$int(sin x+cos x)dx=v, v^2=1-2sin xcos xiffsin xcos x=?$$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 25 at 16:34

























                      answered Nov 24 at 17:49









                      lab bhattacharjee

                      222k15155273




                      222k15155273












                      • Could you expand a Little bit?
                        – Andres Oropeza
                        Nov 24 at 19:21










                      • @AndresOropeza, Please find the updated answer
                        – lab bhattacharjee
                        Nov 25 at 16:34


















                      • Could you expand a Little bit?
                        – Andres Oropeza
                        Nov 24 at 19:21










                      • @AndresOropeza, Please find the updated answer
                        – lab bhattacharjee
                        Nov 25 at 16:34
















                      Could you expand a Little bit?
                      – Andres Oropeza
                      Nov 24 at 19:21




                      Could you expand a Little bit?
                      – Andres Oropeza
                      Nov 24 at 19:21












                      @AndresOropeza, Please find the updated answer
                      – lab bhattacharjee
                      Nov 25 at 16:34




                      @AndresOropeza, Please find the updated answer
                      – lab bhattacharjee
                      Nov 25 at 16:34


















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