Integration of trigonometric function $intfrac{sin(2x)}{sin(x)-cos(x)}dx$












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$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$




My attempt: Firstly, $sin(2x)=2sin(x)cos(x)$.



After that, eliminate the $cos(x)$ seen in both the numerator and denominator to get



$$2intfrac{sin(x)}{tan(x)-1} dx.$$



From here onwards, should I convert $sin(x)$, $tan(x)$ to half-angles and use $tan(x/2)=t$?



But this would be a time consuming method. Any suggestions?










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  • 1




    $begingroup$
    As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
    $endgroup$
    – Yuriy S
    Aug 6 '16 at 11:10


















3












$begingroup$



$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$




My attempt: Firstly, $sin(2x)=2sin(x)cos(x)$.



After that, eliminate the $cos(x)$ seen in both the numerator and denominator to get



$$2intfrac{sin(x)}{tan(x)-1} dx.$$



From here onwards, should I convert $sin(x)$, $tan(x)$ to half-angles and use $tan(x/2)=t$?



But this would be a time consuming method. Any suggestions?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
    $endgroup$
    – Yuriy S
    Aug 6 '16 at 11:10
















3












3








3


2



$begingroup$



$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$




My attempt: Firstly, $sin(2x)=2sin(x)cos(x)$.



After that, eliminate the $cos(x)$ seen in both the numerator and denominator to get



$$2intfrac{sin(x)}{tan(x)-1} dx.$$



From here onwards, should I convert $sin(x)$, $tan(x)$ to half-angles and use $tan(x/2)=t$?



But this would be a time consuming method. Any suggestions?










share|cite|improve this question











$endgroup$





$$intfrac{sin(2x)}{sin(x)-cos(x)}dx$$




My attempt: Firstly, $sin(2x)=2sin(x)cos(x)$.



After that, eliminate the $cos(x)$ seen in both the numerator and denominator to get



$$2intfrac{sin(x)}{tan(x)-1} dx.$$



From here onwards, should I convert $sin(x)$, $tan(x)$ to half-angles and use $tan(x/2)=t$?



But this would be a time consuming method. Any suggestions?







real-analysis integration trigonometry indefinite-integrals






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edited Dec 26 '18 at 23:20









DavidG

2,1121723




2,1121723










asked Aug 6 '16 at 11:07









Archis WelankarArchis Welankar

12k41642




12k41642








  • 1




    $begingroup$
    As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
    $endgroup$
    – Yuriy S
    Aug 6 '16 at 11:10
















  • 1




    $begingroup$
    As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
    $endgroup$
    – Yuriy S
    Aug 6 '16 at 11:10










1




1




$begingroup$
As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
$endgroup$
– Yuriy S
Aug 6 '16 at 11:10






$begingroup$
As I see it, the half-angle tangent substitution, which you mentioned, is a very simple way here. Just try it - the numerator and denominator become very nice
$endgroup$
– Yuriy S
Aug 6 '16 at 11:10












6 Answers
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oldest

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$begingroup$

Hint:
$$intfrac{sin(2x)}{sin(x)-cos(x)}dx = intfrac{2sin(x)cos(x) }{sin(x)-cos(x)}dx = intfrac{sin(x)cos(x)+cos(x)sin(x) }{sin(x)-cos(x)}dx\
=intfrac{sin(x)cos(x)-sin^2(x)+cos(x)sin(x)-cos^2(x)+1 }{sin(x)-cos(x)}dx \=int -sin(x)dx+int cos(x)dx + int frac{1}{sin(x) -cos(x)}dx \= cos(x) + sin(x) +int frac{1}{sin(x) -cos(x)}dx $$



For $int frac{1}{sin(x) -cos(x)}dx$:



Notice that $$int frac{1}{sin(x) -cos(x)}dx = int frac{1}{sqrt{2} sin(x-frac{1}{4} pi ) }dx$$



Let $u = x-frac{1}{4} pi$,



$$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
    $endgroup$
    – Behnam Esmayli
    Aug 6 '16 at 16:44










  • $begingroup$
    @Behnam Nope. there is no sign mistake.
    $endgroup$
    – Zack Ni
    Aug 7 '16 at 1:24



















4












$begingroup$

HINT:



$$intfrac{sin(2x)}{sin(x)-cos(x)}spacetext{d}x=$$





Use $sin(2x)=2sin(x)cos(x)$:





$$2intfrac{sin(x)cos(x)}{sin(x)-cos(x)}spacetext{d}x=$$





Sustitute $u=tanleft(frac{x}{2}right)$ and $text{d}u=frac{xsec^2left(frac{x}{2}right)}{2}spacetext{d}x$:





$$-8intfrac{u(u^2-1)}{(u^2+1)^2(u^2+2u-1)}spacetext{d}u$$



Now, use partial fractions.






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    4












    $begingroup$

    Let $u=x-fracpi4$, then
    $$
    begin{align}
    intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
    &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
    &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
    &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
    &=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
    &=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
    &=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
    &=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
    end{align}
    $$






    share|cite|improve this answer











    $endgroup$





















      3












      $begingroup$

      Let $$I = intfrac{sin 2x}{sin x-cos x}dx = frac{1}{sqrt{2}}intfrac{sin 2x}{sin left(x-frac{pi}{4}right)}dx$$



      Now Put $displaystyle x- frac{pi}{4} = t;,$ Then $dx = dt$



      So $$I = frac{1}{sqrt{2}}intfrac{cos 2t}{sin t}dt = frac{1}{sqrt{2}}int frac{1-2sin^2 t}{sin t}dt$$



      So $$I = frac{1}{sqrt{2}}int csc t dt-sqrt{2}int sin t dt$$



      So $$I = frac{1}{sqrt{2}}ln left|tan frac{t}{2}right|+sqrt{2}cos t+mathcal{C}$$



      So $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
        $endgroup$
        – Ahmed S. Attaalla
        Aug 6 '16 at 12:30








      • 1




        $begingroup$
        There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
        $endgroup$
        – user220382
        Aug 6 '16 at 13:05



















      1












      $begingroup$

      I like trigonometric substitutions so I will try to implement that !



      Use $sin(x)-cos(x)=z$.Find numerator in terms of $z^2$.



      Then $(cos(x)+sin(x))dx=dz$



      $sin(x)-cos(x)=z$
      so,$z^2=1-sin(2x)$



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



      So the integral boils down to



      $$frac{1-z^2}{z(sqrt{2-z^2})}dz$$



      Substitute $z=sqrt{2}sin(y)$
      The integral becomes $$int frac{cos(2y)}{sqrt{2}sin(y)}dy$$
      $$int frac{csc(y)-2sin(y)}{sqrt{2}}dy$$
      which equals $$frac{1}{sqrt{2}}ln|tan(y/2)|+sqrt{2}cos(y)+C$$



      On resubstituting original variables we get $$frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$



      Hurray ! :-)






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        Assuming $$u=x-fracpi4$$
        $$
        begin{align}
        intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
        &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
        &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
        &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
        &=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
        &=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
        &=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
        end{align}
        $$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          There is a mistake in your final answer.Otherwise your method is nice!
          $endgroup$
          – user220382
          Aug 6 '16 at 16:17










        • $begingroup$
          @SanchayanDutta Fixed it
          $endgroup$
          – Aakash Kumar
          Aug 6 '16 at 16:23











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        6 Answers
        6






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        6 Answers
        6






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        0












        $begingroup$

        Hint:
        $$intfrac{sin(2x)}{sin(x)-cos(x)}dx = intfrac{2sin(x)cos(x) }{sin(x)-cos(x)}dx = intfrac{sin(x)cos(x)+cos(x)sin(x) }{sin(x)-cos(x)}dx\
        =intfrac{sin(x)cos(x)-sin^2(x)+cos(x)sin(x)-cos^2(x)+1 }{sin(x)-cos(x)}dx \=int -sin(x)dx+int cos(x)dx + int frac{1}{sin(x) -cos(x)}dx \= cos(x) + sin(x) +int frac{1}{sin(x) -cos(x)}dx $$



        For $int frac{1}{sin(x) -cos(x)}dx$:



        Notice that $$int frac{1}{sin(x) -cos(x)}dx = int frac{1}{sqrt{2} sin(x-frac{1}{4} pi ) }dx$$



        Let $u = x-frac{1}{4} pi$,



        $$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
          $endgroup$
          – Behnam Esmayli
          Aug 6 '16 at 16:44










        • $begingroup$
          @Behnam Nope. there is no sign mistake.
          $endgroup$
          – Zack Ni
          Aug 7 '16 at 1:24
















        0












        $begingroup$

        Hint:
        $$intfrac{sin(2x)}{sin(x)-cos(x)}dx = intfrac{2sin(x)cos(x) }{sin(x)-cos(x)}dx = intfrac{sin(x)cos(x)+cos(x)sin(x) }{sin(x)-cos(x)}dx\
        =intfrac{sin(x)cos(x)-sin^2(x)+cos(x)sin(x)-cos^2(x)+1 }{sin(x)-cos(x)}dx \=int -sin(x)dx+int cos(x)dx + int frac{1}{sin(x) -cos(x)}dx \= cos(x) + sin(x) +int frac{1}{sin(x) -cos(x)}dx $$



        For $int frac{1}{sin(x) -cos(x)}dx$:



        Notice that $$int frac{1}{sin(x) -cos(x)}dx = int frac{1}{sqrt{2} sin(x-frac{1}{4} pi ) }dx$$



        Let $u = x-frac{1}{4} pi$,



        $$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
          $endgroup$
          – Behnam Esmayli
          Aug 6 '16 at 16:44










        • $begingroup$
          @Behnam Nope. there is no sign mistake.
          $endgroup$
          – Zack Ni
          Aug 7 '16 at 1:24














        0












        0








        0





        $begingroup$

        Hint:
        $$intfrac{sin(2x)}{sin(x)-cos(x)}dx = intfrac{2sin(x)cos(x) }{sin(x)-cos(x)}dx = intfrac{sin(x)cos(x)+cos(x)sin(x) }{sin(x)-cos(x)}dx\
        =intfrac{sin(x)cos(x)-sin^2(x)+cos(x)sin(x)-cos^2(x)+1 }{sin(x)-cos(x)}dx \=int -sin(x)dx+int cos(x)dx + int frac{1}{sin(x) -cos(x)}dx \= cos(x) + sin(x) +int frac{1}{sin(x) -cos(x)}dx $$



        For $int frac{1}{sin(x) -cos(x)}dx$:



        Notice that $$int frac{1}{sin(x) -cos(x)}dx = int frac{1}{sqrt{2} sin(x-frac{1}{4} pi ) }dx$$



        Let $u = x-frac{1}{4} pi$,



        $$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$






        share|cite|improve this answer











        $endgroup$



        Hint:
        $$intfrac{sin(2x)}{sin(x)-cos(x)}dx = intfrac{2sin(x)cos(x) }{sin(x)-cos(x)}dx = intfrac{sin(x)cos(x)+cos(x)sin(x) }{sin(x)-cos(x)}dx\
        =intfrac{sin(x)cos(x)-sin^2(x)+cos(x)sin(x)-cos^2(x)+1 }{sin(x)-cos(x)}dx \=int -sin(x)dx+int cos(x)dx + int frac{1}{sin(x) -cos(x)}dx \= cos(x) + sin(x) +int frac{1}{sin(x) -cos(x)}dx $$



        For $int frac{1}{sin(x) -cos(x)}dx$:



        Notice that $$int frac{1}{sin(x) -cos(x)}dx = int frac{1}{sqrt{2} sin(x-frac{1}{4} pi ) }dx$$



        Let $u = x-frac{1}{4} pi$,



        $$ int frac{1}{sqrt{2} sinleft(x-frac{1}{4} pi right) }dx = int frac{1}{sqrt{2} } csc(u)du $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 27 '18 at 0:01









        DavidG

        2,1121723




        2,1121723










        answered Aug 6 '16 at 12:08









        Zack NiZack Ni

        3,466729




        3,466729












        • $begingroup$
          There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
          $endgroup$
          – Behnam Esmayli
          Aug 6 '16 at 16:44










        • $begingroup$
          @Behnam Nope. there is no sign mistake.
          $endgroup$
          – Zack Ni
          Aug 7 '16 at 1:24


















        • $begingroup$
          There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
          $endgroup$
          – Behnam Esmayli
          Aug 6 '16 at 16:44










        • $begingroup$
          @Behnam Nope. there is no sign mistake.
          $endgroup$
          – Zack Ni
          Aug 7 '16 at 1:24
















        $begingroup$
        There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
        $endgroup$
        – Behnam Esmayli
        Aug 6 '16 at 16:44




        $begingroup$
        There is a shortcut: write $sin (2x)=1-(sin (x) - cos (x))^2$. Is there a mistake in your sign? Shouln't it be $-sin (x)?$
        $endgroup$
        – Behnam Esmayli
        Aug 6 '16 at 16:44












        $begingroup$
        @Behnam Nope. there is no sign mistake.
        $endgroup$
        – Zack Ni
        Aug 7 '16 at 1:24




        $begingroup$
        @Behnam Nope. there is no sign mistake.
        $endgroup$
        – Zack Ni
        Aug 7 '16 at 1:24











        4












        $begingroup$

        HINT:



        $$intfrac{sin(2x)}{sin(x)-cos(x)}spacetext{d}x=$$





        Use $sin(2x)=2sin(x)cos(x)$:





        $$2intfrac{sin(x)cos(x)}{sin(x)-cos(x)}spacetext{d}x=$$





        Sustitute $u=tanleft(frac{x}{2}right)$ and $text{d}u=frac{xsec^2left(frac{x}{2}right)}{2}spacetext{d}x$:





        $$-8intfrac{u(u^2-1)}{(u^2+1)^2(u^2+2u-1)}spacetext{d}u$$



        Now, use partial fractions.






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          HINT:



          $$intfrac{sin(2x)}{sin(x)-cos(x)}spacetext{d}x=$$





          Use $sin(2x)=2sin(x)cos(x)$:





          $$2intfrac{sin(x)cos(x)}{sin(x)-cos(x)}spacetext{d}x=$$





          Sustitute $u=tanleft(frac{x}{2}right)$ and $text{d}u=frac{xsec^2left(frac{x}{2}right)}{2}spacetext{d}x$:





          $$-8intfrac{u(u^2-1)}{(u^2+1)^2(u^2+2u-1)}spacetext{d}u$$



          Now, use partial fractions.






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            HINT:



            $$intfrac{sin(2x)}{sin(x)-cos(x)}spacetext{d}x=$$





            Use $sin(2x)=2sin(x)cos(x)$:





            $$2intfrac{sin(x)cos(x)}{sin(x)-cos(x)}spacetext{d}x=$$





            Sustitute $u=tanleft(frac{x}{2}right)$ and $text{d}u=frac{xsec^2left(frac{x}{2}right)}{2}spacetext{d}x$:





            $$-8intfrac{u(u^2-1)}{(u^2+1)^2(u^2+2u-1)}spacetext{d}u$$



            Now, use partial fractions.






            share|cite|improve this answer









            $endgroup$



            HINT:



            $$intfrac{sin(2x)}{sin(x)-cos(x)}spacetext{d}x=$$





            Use $sin(2x)=2sin(x)cos(x)$:





            $$2intfrac{sin(x)cos(x)}{sin(x)-cos(x)}spacetext{d}x=$$





            Sustitute $u=tanleft(frac{x}{2}right)$ and $text{d}u=frac{xsec^2left(frac{x}{2}right)}{2}spacetext{d}x$:





            $$-8intfrac{u(u^2-1)}{(u^2+1)^2(u^2+2u-1)}spacetext{d}u$$



            Now, use partial fractions.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 6 '16 at 11:13









            JanJan

            21.8k31240




            21.8k31240























                4












                $begingroup$

                Let $u=x-fracpi4$, then
                $$
                begin{align}
                intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
                &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
                &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
                &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
                &=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
                &=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
                &=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
                &=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
                end{align}
                $$






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  Let $u=x-fracpi4$, then
                  $$
                  begin{align}
                  intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
                  &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
                  &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
                  &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
                  &=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
                  &=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
                  &=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
                  &=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
                  end{align}
                  $$






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Let $u=x-fracpi4$, then
                    $$
                    begin{align}
                    intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
                    &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
                    &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
                    &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
                    &=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
                    &=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
                    &=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
                    &=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
                    end{align}
                    $$






                    share|cite|improve this answer











                    $endgroup$



                    Let $u=x-fracpi4$, then
                    $$
                    begin{align}
                    intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
                    &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
                    &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
                    &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
                    &=frac1{sqrt2}intfrac{sin(u)}{1-cos^2(u)},mathrm{d}u-sqrt2intsin(u),mathrm{d}u\
                    &=sqrt2cos(u)-frac1{2sqrt2}intleft(frac1{1-cos(u)}+frac1{1+cos(u)}right),mathrm{d}cos(u)\
                    &=sqrt2cos(u)+frac1{2sqrt2}logleft(frac{1-cos(u)}{1+cos(u)}right)+C\
                    &=sqrt2cos(u)+frac1{sqrt2}log(tan(u/2))+C
                    end{align}
                    $$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Aug 6 '16 at 13:37

























                    answered Aug 6 '16 at 13:30









                    robjohnrobjohn

                    268k27308633




                    268k27308633























                        3












                        $begingroup$

                        Let $$I = intfrac{sin 2x}{sin x-cos x}dx = frac{1}{sqrt{2}}intfrac{sin 2x}{sin left(x-frac{pi}{4}right)}dx$$



                        Now Put $displaystyle x- frac{pi}{4} = t;,$ Then $dx = dt$



                        So $$I = frac{1}{sqrt{2}}intfrac{cos 2t}{sin t}dt = frac{1}{sqrt{2}}int frac{1-2sin^2 t}{sin t}dt$$



                        So $$I = frac{1}{sqrt{2}}int csc t dt-sqrt{2}int sin t dt$$



                        So $$I = frac{1}{sqrt{2}}ln left|tan frac{t}{2}right|+sqrt{2}cos t+mathcal{C}$$



                        So $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$






                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
                          $endgroup$
                          – Ahmed S. Attaalla
                          Aug 6 '16 at 12:30








                        • 1




                          $begingroup$
                          There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
                          $endgroup$
                          – user220382
                          Aug 6 '16 at 13:05
















                        3












                        $begingroup$

                        Let $$I = intfrac{sin 2x}{sin x-cos x}dx = frac{1}{sqrt{2}}intfrac{sin 2x}{sin left(x-frac{pi}{4}right)}dx$$



                        Now Put $displaystyle x- frac{pi}{4} = t;,$ Then $dx = dt$



                        So $$I = frac{1}{sqrt{2}}intfrac{cos 2t}{sin t}dt = frac{1}{sqrt{2}}int frac{1-2sin^2 t}{sin t}dt$$



                        So $$I = frac{1}{sqrt{2}}int csc t dt-sqrt{2}int sin t dt$$



                        So $$I = frac{1}{sqrt{2}}ln left|tan frac{t}{2}right|+sqrt{2}cos t+mathcal{C}$$



                        So $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$






                        share|cite|improve this answer











                        $endgroup$









                        • 1




                          $begingroup$
                          Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
                          $endgroup$
                          – Ahmed S. Attaalla
                          Aug 6 '16 at 12:30








                        • 1




                          $begingroup$
                          There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
                          $endgroup$
                          – user220382
                          Aug 6 '16 at 13:05














                        3












                        3








                        3





                        $begingroup$

                        Let $$I = intfrac{sin 2x}{sin x-cos x}dx = frac{1}{sqrt{2}}intfrac{sin 2x}{sin left(x-frac{pi}{4}right)}dx$$



                        Now Put $displaystyle x- frac{pi}{4} = t;,$ Then $dx = dt$



                        So $$I = frac{1}{sqrt{2}}intfrac{cos 2t}{sin t}dt = frac{1}{sqrt{2}}int frac{1-2sin^2 t}{sin t}dt$$



                        So $$I = frac{1}{sqrt{2}}int csc t dt-sqrt{2}int sin t dt$$



                        So $$I = frac{1}{sqrt{2}}ln left|tan frac{t}{2}right|+sqrt{2}cos t+mathcal{C}$$



                        So $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$






                        share|cite|improve this answer











                        $endgroup$



                        Let $$I = intfrac{sin 2x}{sin x-cos x}dx = frac{1}{sqrt{2}}intfrac{sin 2x}{sin left(x-frac{pi}{4}right)}dx$$



                        Now Put $displaystyle x- frac{pi}{4} = t;,$ Then $dx = dt$



                        So $$I = frac{1}{sqrt{2}}intfrac{cos 2t}{sin t}dt = frac{1}{sqrt{2}}int frac{1-2sin^2 t}{sin t}dt$$



                        So $$I = frac{1}{sqrt{2}}int csc t dt-sqrt{2}int sin t dt$$



                        So $$I = frac{1}{sqrt{2}}ln left|tan frac{t}{2}right|+sqrt{2}cos t+mathcal{C}$$



                        So $$I = frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Aug 6 '16 at 13:46

























                        answered Aug 6 '16 at 12:09









                        juantheronjuantheron

                        34.3k1147142




                        34.3k1147142








                        • 1




                          $begingroup$
                          Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
                          $endgroup$
                          – Ahmed S. Attaalla
                          Aug 6 '16 at 12:30








                        • 1




                          $begingroup$
                          There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
                          $endgroup$
                          – user220382
                          Aug 6 '16 at 13:05














                        • 1




                          $begingroup$
                          Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
                          $endgroup$
                          – Ahmed S. Attaalla
                          Aug 6 '16 at 12:30








                        • 1




                          $begingroup$
                          There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
                          $endgroup$
                          – user220382
                          Aug 6 '16 at 13:05








                        1




                        1




                        $begingroup$
                        Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
                        $endgroup$
                        – Ahmed S. Attaalla
                        Aug 6 '16 at 12:30






                        $begingroup$
                        Clever approach. How did you know that $sin x-cos x=sqrt{2}sin(x-frac{pi}{4})$. Did you use the sum identities and equate coefficients?
                        $endgroup$
                        – Ahmed S. Attaalla
                        Aug 6 '16 at 12:30






                        1




                        1




                        $begingroup$
                        There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
                        $endgroup$
                        – user220382
                        Aug 6 '16 at 13:05




                        $begingroup$
                        There will a $sqrt(2)$ before $cos(x-pi/4)$ in last step.
                        $endgroup$
                        – user220382
                        Aug 6 '16 at 13:05











                        1












                        $begingroup$

                        I like trigonometric substitutions so I will try to implement that !



                        Use $sin(x)-cos(x)=z$.Find numerator in terms of $z^2$.



                        Then $(cos(x)+sin(x))dx=dz$



                        $sin(x)-cos(x)=z$
                        so,$z^2=1-sin(2x)$



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



                        So the integral boils down to



                        $$frac{1-z^2}{z(sqrt{2-z^2})}dz$$



                        Substitute $z=sqrt{2}sin(y)$
                        The integral becomes $$int frac{cos(2y)}{sqrt{2}sin(y)}dy$$
                        $$int frac{csc(y)-2sin(y)}{sqrt{2}}dy$$
                        which equals $$frac{1}{sqrt{2}}ln|tan(y/2)|+sqrt{2}cos(y)+C$$



                        On resubstituting original variables we get $$frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$



                        Hurray ! :-)






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          I like trigonometric substitutions so I will try to implement that !



                          Use $sin(x)-cos(x)=z$.Find numerator in terms of $z^2$.



                          Then $(cos(x)+sin(x))dx=dz$



                          $sin(x)-cos(x)=z$
                          so,$z^2=1-sin(2x)$



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



                          So the integral boils down to



                          $$frac{1-z^2}{z(sqrt{2-z^2})}dz$$



                          Substitute $z=sqrt{2}sin(y)$
                          The integral becomes $$int frac{cos(2y)}{sqrt{2}sin(y)}dy$$
                          $$int frac{csc(y)-2sin(y)}{sqrt{2}}dy$$
                          which equals $$frac{1}{sqrt{2}}ln|tan(y/2)|+sqrt{2}cos(y)+C$$



                          On resubstituting original variables we get $$frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$



                          Hurray ! :-)






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            I like trigonometric substitutions so I will try to implement that !



                            Use $sin(x)-cos(x)=z$.Find numerator in terms of $z^2$.



                            Then $(cos(x)+sin(x))dx=dz$



                            $sin(x)-cos(x)=z$
                            so,$z^2=1-sin(2x)$



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



                            So the integral boils down to



                            $$frac{1-z^2}{z(sqrt{2-z^2})}dz$$



                            Substitute $z=sqrt{2}sin(y)$
                            The integral becomes $$int frac{cos(2y)}{sqrt{2}sin(y)}dy$$
                            $$int frac{csc(y)-2sin(y)}{sqrt{2}}dy$$
                            which equals $$frac{1}{sqrt{2}}ln|tan(y/2)|+sqrt{2}cos(y)+C$$



                            On resubstituting original variables we get $$frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$



                            Hurray ! :-)






                            share|cite|improve this answer











                            $endgroup$



                            I like trigonometric substitutions so I will try to implement that !



                            Use $sin(x)-cos(x)=z$.Find numerator in terms of $z^2$.



                            Then $(cos(x)+sin(x))dx=dz$



                            $sin(x)-cos(x)=z$
                            so,$z^2=1-sin(2x)$



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



                            So the integral boils down to



                            $$frac{1-z^2}{z(sqrt{2-z^2})}dz$$



                            Substitute $z=sqrt{2}sin(y)$
                            The integral becomes $$int frac{cos(2y)}{sqrt{2}sin(y)}dy$$
                            $$int frac{csc(y)-2sin(y)}{sqrt{2}}dy$$
                            which equals $$frac{1}{sqrt{2}}ln|tan(y/2)|+sqrt{2}cos(y)+C$$



                            On resubstituting original variables we get $$frac{1}{sqrt{2}}ln left|tan left(frac{x-frac{pi}{4}}{2}right)right|+sqrt{2}cos left(x-frac{pi}{4}right)+mathcal{C}$$



                            Hurray ! :-)







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Aug 6 '16 at 13:06

























                            answered Aug 6 '16 at 12:09







                            user220382






























                                1












                                $begingroup$

                                Assuming $$u=x-fracpi4$$
                                $$
                                begin{align}
                                intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
                                &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
                                &=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
                                &=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
                                end{align}
                                $$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  There is a mistake in your final answer.Otherwise your method is nice!
                                  $endgroup$
                                  – user220382
                                  Aug 6 '16 at 16:17










                                • $begingroup$
                                  @SanchayanDutta Fixed it
                                  $endgroup$
                                  – Aakash Kumar
                                  Aug 6 '16 at 16:23
















                                1












                                $begingroup$

                                Assuming $$u=x-fracpi4$$
                                $$
                                begin{align}
                                intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
                                &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
                                &=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
                                &=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
                                end{align}
                                $$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  There is a mistake in your final answer.Otherwise your method is nice!
                                  $endgroup$
                                  – user220382
                                  Aug 6 '16 at 16:17










                                • $begingroup$
                                  @SanchayanDutta Fixed it
                                  $endgroup$
                                  – Aakash Kumar
                                  Aug 6 '16 at 16:23














                                1












                                1








                                1





                                $begingroup$

                                Assuming $$u=x-fracpi4$$
                                $$
                                begin{align}
                                intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
                                &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
                                &=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
                                &=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
                                end{align}
                                $$






                                share|cite|improve this answer











                                $endgroup$



                                Assuming $$u=x-fracpi4$$
                                $$
                                begin{align}
                                intfrac{sin(2x)}{sin(x)-cos(x)},mathrm{d}x
                                &=intfrac{sinleft(2u+fracpi2right)}{sqrt2sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intfrac{cosleft(2uright)}{sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intfrac{1-2sin^2(u)}{sin(u)},mathrm{d}u\
                                &=frac1{sqrt2}intleft(csc u -2sin uright),mathrm{d}u\
                                &=frac1{sqrt2}int csc u text{du} -2sin u text{du} \
                                &=frac1{sqrt2}left(log left|tan frac{u}{2}right| +2cos uright) \
                                end{align}
                                $$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Aug 6 '16 at 16:18

























                                answered Aug 6 '16 at 15:43









                                Aakash KumarAakash Kumar

                                2,5211519




                                2,5211519












                                • $begingroup$
                                  There is a mistake in your final answer.Otherwise your method is nice!
                                  $endgroup$
                                  – user220382
                                  Aug 6 '16 at 16:17










                                • $begingroup$
                                  @SanchayanDutta Fixed it
                                  $endgroup$
                                  – Aakash Kumar
                                  Aug 6 '16 at 16:23


















                                • $begingroup$
                                  There is a mistake in your final answer.Otherwise your method is nice!
                                  $endgroup$
                                  – user220382
                                  Aug 6 '16 at 16:17










                                • $begingroup$
                                  @SanchayanDutta Fixed it
                                  $endgroup$
                                  – Aakash Kumar
                                  Aug 6 '16 at 16:23
















                                $begingroup$
                                There is a mistake in your final answer.Otherwise your method is nice!
                                $endgroup$
                                – user220382
                                Aug 6 '16 at 16:17




                                $begingroup$
                                There is a mistake in your final answer.Otherwise your method is nice!
                                $endgroup$
                                – user220382
                                Aug 6 '16 at 16:17












                                $begingroup$
                                @SanchayanDutta Fixed it
                                $endgroup$
                                – Aakash Kumar
                                Aug 6 '16 at 16:23




                                $begingroup$
                                @SanchayanDutta Fixed it
                                $endgroup$
                                – Aakash Kumar
                                Aug 6 '16 at 16:23


















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