Evalutating an indefinit integral

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How can I evaluate the following integral?
$$
intfraccosx1+sin2x dx
$$




I tried the following way, but I was not able to proceed further:
$$
begingather
I&=intfraccosxleft(sinx+cosxright)^2 dx\
&= intfracsecxleft(1+tanxright)^2 dx
endgather
$$







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  • Hope you'll satisty with the solution.
    – Arpit Yadav
    Aug 12 at 3:01














up vote
0
down vote

favorite













How can I evaluate the following integral?
$$
intfraccosx1+sin2x dx
$$




I tried the following way, but I was not able to proceed further:
$$
begingather
I&=intfraccosxleft(sinx+cosxright)^2 dx\
&= intfracsecxleft(1+tanxright)^2 dx
endgather
$$







share|cite|improve this question






















  • Hope you'll satisty with the solution.
    – Arpit Yadav
    Aug 12 at 3:01












up vote
0
down vote

favorite









up vote
0
down vote

favorite












How can I evaluate the following integral?
$$
intfraccosx1+sin2x dx
$$




I tried the following way, but I was not able to proceed further:
$$
begingather
I&=intfraccosxleft(sinx+cosxright)^2 dx\
&= intfracsecxleft(1+tanxright)^2 dx
endgather
$$







share|cite|improve this question















How can I evaluate the following integral?
$$
intfraccosx1+sin2x dx
$$




I tried the following way, but I was not able to proceed further:
$$
begingather
I&=intfraccosxleft(sinx+cosxright)^2 dx\
&= intfracsecxleft(1+tanxright)^2 dx
endgather
$$









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edited Aug 12 at 1:42









Math Lover

12.5k21232




12.5k21232










asked Aug 12 at 0:17









Hussien Mohamed

634112




634112











  • Hope you'll satisty with the solution.
    – Arpit Yadav
    Aug 12 at 3:01
















  • Hope you'll satisty with the solution.
    – Arpit Yadav
    Aug 12 at 3:01















Hope you'll satisty with the solution.
– Arpit Yadav
Aug 12 at 3:01




Hope you'll satisty with the solution.
– Arpit Yadav
Aug 12 at 3:01










2 Answers
2






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oldest

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up vote
2
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Hint:



Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$






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    Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
    $$I=I_1+I_2$$
    $$I_1=int frac1cos x+sin xdx$$
    $$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
    $$I_1=intfracsec(x-pi/4)sqrt2dx$$
    $$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
    Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
    $$sin x+cos x=timplies(cos x-sin x)dx=dt$$
    $$I_2=intfrac 1t^2space dt$$
    $$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$






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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      up vote
      2
      down vote













      Hint:



      Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$






      share|cite|improve this answer
























        up vote
        2
        down vote













        Hint:



        Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$






        share|cite|improve this answer






















          up vote
          2
          down vote










          up vote
          2
          down vote









          Hint:



          Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$






          share|cite|improve this answer












          Hint:



          Write numerator as $$2cos x=cos x+sin x +(cos x-sin x)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 12 at 1:34









          lab bhattacharjee

          215k14152264




          215k14152264




















              up vote
              0
              down vote













              Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
              $$I=I_1+I_2$$
              $$I_1=int frac1cos x+sin xdx$$
              $$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
              $$I_1=intfracsec(x-pi/4)sqrt2dx$$
              $$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
              Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
              $$sin x+cos x=timplies(cos x-sin x)dx=dt$$
              $$I_2=intfrac 1t^2space dt$$
              $$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$






              share|cite|improve this answer


























                up vote
                0
                down vote













                Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
                $$I=I_1+I_2$$
                $$I_1=int frac1cos x+sin xdx$$
                $$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
                $$I_1=intfracsec(x-pi/4)sqrt2dx$$
                $$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
                Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
                $$sin x+cos x=timplies(cos x-sin x)dx=dt$$
                $$I_2=intfrac 1t^2space dt$$
                $$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$






                share|cite|improve this answer
























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
                  $$I=I_1+I_2$$
                  $$I_1=int frac1cos x+sin xdx$$
                  $$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
                  $$I_1=intfracsec(x-pi/4)sqrt2dx$$
                  $$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
                  Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
                  $$sin x+cos x=timplies(cos x-sin x)dx=dt$$
                  $$I_2=intfrac 1t^2space dt$$
                  $$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$






                  share|cite|improve this answer














                  Using lab bhattacharjee result,$$I=int fraccos x+ sin x(cos x+ sin x)^2dxspace +int fraccos x-sin x(cos x+sin x)^2dx$$
                  $$I=I_1+I_2$$
                  $$I_1=int frac1cos x+sin xdx$$
                  $$I_1=intfrac1sqrt2(cos xcdotcosfracpi4+sin xcdot sinfracpi4)dx$$
                  $$I_1=intfracsec(x-pi/4)sqrt2dx$$
                  $$I_1=frac1sqrt2cdotln[(sec(x-pi/4)+tan(x-pi/4)]+C_1$$
                  Now$$I_2=int fraccos x-sin x(cos x+sin x)^2dx$$
                  $$sin x+cos x=timplies(cos x-sin x)dx=dt$$
                  $$I_2=intfrac 1t^2space dt$$
                  $$I_2=-frac1t+C_2=-frac1sin x+cos x+C_2$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 12 at 3:00

























                  answered Aug 12 at 2:54









                  Arpit Yadav

                  331215




                  331215






















                       

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