Evaluating $int frac {3cos x}{(2sin x-5cos x)},dx$











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$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$



I've been thinking and trying to work this out in quite a few ways:



1)Taking conjugate which actually complicates it further



2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.



3)Dividing throughout by $cos^2 x$



I find none of these methods to be effective. Please advice.






calculus integration







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  • math.stackexchange.com/questions/3006097/…
    – lab bhattacharjee
    Nov 22 at 14:55















up vote
0
down vote

favorite












$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$



I've been thinking and trying to work this out in quite a few ways:



1)Taking conjugate which actually complicates it further



2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.



3)Dividing throughout by $cos^2 x$



I find none of these methods to be effective. Please advice.






calculus integration







share|cite|improve this question
























  • math.stackexchange.com/questions/3006097/…
    – lab bhattacharjee
    Nov 22 at 14:55













up vote
0
down vote

favorite









up vote
0
down vote

favorite











$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$



I've been thinking and trying to work this out in quite a few ways:



1)Taking conjugate which actually complicates it further



2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.



3)Dividing throughout by $cos^2 x$



I find none of these methods to be effective. Please advice.






calculus integration







share|cite|improve this question















$$int dfrac {3cos x}{(2sin x-5cos x)},dx$$



I've been thinking and trying to work this out in quite a few ways:



1)Taking conjugate which actually complicates it further



2)Using half angle formula to convert the expression in terms of tan but I get 2 terms in the integration of which I'm unable to integrate 1 term because I can't seem to be able to make a suitable substitution.



3)Dividing throughout by $cos^2 x$



I find none of these methods to be effective. Please advice.






calculus integration




calculus integration






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edited Nov 22 at 14:58









StubbornAtom

5,06011138




5,06011138










asked Nov 22 at 14:48









MadCap

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213












  • math.stackexchange.com/questions/3006097/…
    – lab bhattacharjee
    Nov 22 at 14:55


















  • math.stackexchange.com/questions/3006097/…
    – lab bhattacharjee
    Nov 22 at 14:55
















math.stackexchange.com/questions/3006097/…
– lab bhattacharjee
Nov 22 at 14:55




math.stackexchange.com/questions/3006097/…
– lab bhattacharjee
Nov 22 at 14:55










3 Answers
3






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up vote
3
down vote













Hint. A standard trick is to consider
$$
I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
$$ then one may observe that
$$
I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
$$Hope you can take it from here.






share|cite|improve this answer




























    up vote
    1
    down vote













    Use:



    $3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $



    Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:



    $lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$



    So we have:



    $$int frac {3cos x}{(2sin x-5cos x)},dx
    \ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$






    share|cite|improve this answer






























      up vote
      0
      down vote













      Hint:



      By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote













        Hint. A standard trick is to consider
        $$
        I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
        $$         then one may observe that
        $$
        I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
        $$        Hope you can take it from here.






        share|cite|improve this answer

























          up vote
          3
          down vote













          Hint. A standard trick is to consider
          $$
          I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
          $$           then one may observe that
          $$
          I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
          $$          Hope you can take it from here.






          share|cite|improve this answer























            up vote
            3
            down vote










            up vote
            3
            down vote









            Hint. A standard trick is to consider
            $$
            I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
            $$             then one may observe that
            $$
            I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
            $$            Hope you can take it from here.






            share|cite|improve this answer












            Hint. A standard trick is to consider
            $$
            I=intfrac{sin x}{sin x-5cos x}dx,quad J=intfrac{cos x}{sin x-5cos x}dx
            $$             then one may observe that
            $$
            I-5J=int dx=color{red}?,qquad 5I+J=intfrac{(sin x-5cos x)'}{sin x-5cos x}dx=color{red}?
            $$            Hope you can take it from here.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 at 14:55









            Olivier Oloa

            107k17175293




            107k17175293






















                up vote
                1
                down vote













                Use:



                $3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $



                Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:



                $lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$



                So we have:



                $$int frac {3cos x}{(2sin x-5cos x)},dx
                \ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$






                share|cite|improve this answer



























                  up vote
                  1
                  down vote













                  Use:



                  $3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $



                  Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:



                  $lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$



                  So we have:



                  $$int frac {3cos x}{(2sin x-5cos x)},dx
                  \ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Use:



                    $3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $



                    Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:



                    $lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$



                    So we have:



                    $$int frac {3cos x}{(2sin x-5cos x)},dx
                    \ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$






                    share|cite|improve this answer














                    Use:



                    $3 cos x = lambda (2sin x - 5cos x) + mu(2cos x + 5 sin x) $



                    Putting $x = 0$ and $x = pi/2 $ and solving the two equations we get:



                    $lambda = -dfrac {15}{29} $ and $mu = dfrac 6{29}$



                    So we have:



                    $$int frac {3cos x}{(2sin x-5cos x)},dx
                    \ = int dfrac{lambda(2sin x - 5 cos x)+ mu(2cos x + 5 sin x)}{2sin x - 5 cos x} \ = int lambda dx + muint dfrac{2cos x+ 5sin x}{2 sin x - 5 cos x}dx \ =lambda x + mu ln (2sin x - 5cos x) + c = -dfrac{15}{29} x + dfrac{6}{29} ln (2sin x - 5cos x) + c$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 22 at 15:07

























                    answered Nov 22 at 14:56









                    Abcd

                    2,93221131




                    2,93221131






















                        up vote
                        0
                        down vote













                        Hint:



                        By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Hint:



                          By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Hint:



                            By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.






                            share|cite|improve this answer












                            Hint:



                            By shifting the argument, you can turn the denominator to $sqrt{29}cos x$. Then the numerator becomes of the form $acos x+bsin x$, and the fraction is easy to integrate.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 at 15:04









                            Yves Daoust

                            123k668219




                            123k668219






























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