Evaluating $int frac{cos x}{sin^3x+sin x}dx$












2












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Given the function $$g(x)=frac{cos x}{sin^3x+sin x}$$, by letting $u=sin x$, show that $$int g(x) dx=intleft(frac{A}{u}+frac{Bu+C}{u^2+1}right)du$$ where $A,B$ and $C$ are constants. Hence, find $A,B$ and $C$. Hence, solve $int g(x)dx$.



My attempt,
$$g(x)=frac{cos x}{sin^2 x(sin x+1)}$$



$$=frac{sqrt{1-u^2}}{u^2(u+1)}$$



I'm stuck here.






calculus integration indefinite-integrals







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  • $begingroup$
    Interesting that the recommended solution uses partial fractions rather than a second trig substitution.
    $endgroup$
    – Doug M
    Dec 13 '18 at 16:47
















2












$begingroup$


Given the function $$g(x)=frac{cos x}{sin^3x+sin x}$$, by letting $u=sin x$, show that $$int g(x) dx=intleft(frac{A}{u}+frac{Bu+C}{u^2+1}right)du$$ where $A,B$ and $C$ are constants. Hence, find $A,B$ and $C$. Hence, solve $int g(x)dx$.



My attempt,
$$g(x)=frac{cos x}{sin^2 x(sin x+1)}$$



$$=frac{sqrt{1-u^2}}{u^2(u+1)}$$



I'm stuck here.






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Interesting that the recommended solution uses partial fractions rather than a second trig substitution.
    $endgroup$
    – Doug M
    Dec 13 '18 at 16:47














2












2








2





$begingroup$


Given the function $$g(x)=frac{cos x}{sin^3x+sin x}$$, by letting $u=sin x$, show that $$int g(x) dx=intleft(frac{A}{u}+frac{Bu+C}{u^2+1}right)du$$ where $A,B$ and $C$ are constants. Hence, find $A,B$ and $C$. Hence, solve $int g(x)dx$.



My attempt,
$$g(x)=frac{cos x}{sin^2 x(sin x+1)}$$



$$=frac{sqrt{1-u^2}}{u^2(u+1)}$$



I'm stuck here.






calculus integration indefinite-integrals







share|cite|improve this question











$endgroup$




Given the function $$g(x)=frac{cos x}{sin^3x+sin x}$$, by letting $u=sin x$, show that $$int g(x) dx=intleft(frac{A}{u}+frac{Bu+C}{u^2+1}right)du$$ where $A,B$ and $C$ are constants. Hence, find $A,B$ and $C$. Hence, solve $int g(x)dx$.



My attempt,
$$g(x)=frac{cos x}{sin^2 x(sin x+1)}$$



$$=frac{sqrt{1-u^2}}{u^2(u+1)}$$



I'm stuck here.






calculus integration indefinite-integrals




calculus integration indefinite-integrals






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edited Dec 14 '18 at 12:35









AryanSonwatikar

471114




471114










asked Dec 13 '18 at 16:12









MathxxMathxx

3,41811444




3,41811444












  • $begingroup$
    Interesting that the recommended solution uses partial fractions rather than a second trig substitution.
    $endgroup$
    – Doug M
    Dec 13 '18 at 16:47


















  • $begingroup$
    Interesting that the recommended solution uses partial fractions rather than a second trig substitution.
    $endgroup$
    – Doug M
    Dec 13 '18 at 16:47
















$begingroup$
Interesting that the recommended solution uses partial fractions rather than a second trig substitution.
$endgroup$
– Doug M
Dec 13 '18 at 16:47




$begingroup$
Interesting that the recommended solution uses partial fractions rather than a second trig substitution.
$endgroup$
– Doug M
Dec 13 '18 at 16:47










2 Answers
2






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7












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If you do $u=sin x$, then you must also do $mathrm du=cos x,mathrm dx$. So$$intfrac{cos x}{sin^3x+sin x},mathrm dx$$becomes$$intfrac1{u^3+u},mathrm du=intfrac1{u(u^2+1)},mathrm du.$$Can you take it from here?






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Hint:



    $sin x=uimplies $



    $$int g(x) dx=intdfrac{du}{u(u^2+1)}$$



    Now using Partial Fraction Decomposition let $dfrac1{u(u^2+1)}=dfrac Au+dfrac{Bu+C}{u^2+1}$



    $implies1=A(u^2+1)+u(Bu+C)=u^2(A+B)+Cu+A$



    Comparing the coefficients of $u,u^2$ and the constants



    $implies C=0,A=1,A+B=0iff B=-A=?$



    Can you take it from here?






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

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      7












      $begingroup$

      If you do $u=sin x$, then you must also do $mathrm du=cos x,mathrm dx$. So$$intfrac{cos x}{sin^3x+sin x},mathrm dx$$becomes$$intfrac1{u^3+u},mathrm du=intfrac1{u(u^2+1)},mathrm du.$$Can you take it from here?






      share|cite|improve this answer











      $endgroup$


















        7












        $begingroup$

        If you do $u=sin x$, then you must also do $mathrm du=cos x,mathrm dx$. So$$intfrac{cos x}{sin^3x+sin x},mathrm dx$$becomes$$intfrac1{u^3+u},mathrm du=intfrac1{u(u^2+1)},mathrm du.$$Can you take it from here?






        share|cite|improve this answer











        $endgroup$
















          7












          7








          7





          $begingroup$

          If you do $u=sin x$, then you must also do $mathrm du=cos x,mathrm dx$. So$$intfrac{cos x}{sin^3x+sin x},mathrm dx$$becomes$$intfrac1{u^3+u},mathrm du=intfrac1{u(u^2+1)},mathrm du.$$Can you take it from here?






          share|cite|improve this answer











          $endgroup$



          If you do $u=sin x$, then you must also do $mathrm du=cos x,mathrm dx$. So$$intfrac{cos x}{sin^3x+sin x},mathrm dx$$becomes$$intfrac1{u^3+u},mathrm du=intfrac1{u(u^2+1)},mathrm du.$$Can you take it from here?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 16:25









          Shubham Johri

          5,525818




          5,525818










          answered Dec 13 '18 at 16:16









          José Carlos SantosJosé Carlos Santos

          173k23133241




          173k23133241























              3












              $begingroup$

              Hint:



              $sin x=uimplies $



              $$int g(x) dx=intdfrac{du}{u(u^2+1)}$$



              Now using Partial Fraction Decomposition let $dfrac1{u(u^2+1)}=dfrac Au+dfrac{Bu+C}{u^2+1}$



              $implies1=A(u^2+1)+u(Bu+C)=u^2(A+B)+Cu+A$



              Comparing the coefficients of $u,u^2$ and the constants



              $implies C=0,A=1,A+B=0iff B=-A=?$



              Can you take it from here?






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Hint:



                $sin x=uimplies $



                $$int g(x) dx=intdfrac{du}{u(u^2+1)}$$



                Now using Partial Fraction Decomposition let $dfrac1{u(u^2+1)}=dfrac Au+dfrac{Bu+C}{u^2+1}$



                $implies1=A(u^2+1)+u(Bu+C)=u^2(A+B)+Cu+A$



                Comparing the coefficients of $u,u^2$ and the constants



                $implies C=0,A=1,A+B=0iff B=-A=?$



                Can you take it from here?






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Hint:



                  $sin x=uimplies $



                  $$int g(x) dx=intdfrac{du}{u(u^2+1)}$$



                  Now using Partial Fraction Decomposition let $dfrac1{u(u^2+1)}=dfrac Au+dfrac{Bu+C}{u^2+1}$



                  $implies1=A(u^2+1)+u(Bu+C)=u^2(A+B)+Cu+A$



                  Comparing the coefficients of $u,u^2$ and the constants



                  $implies C=0,A=1,A+B=0iff B=-A=?$



                  Can you take it from here?






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  $sin x=uimplies $



                  $$int g(x) dx=intdfrac{du}{u(u^2+1)}$$



                  Now using Partial Fraction Decomposition let $dfrac1{u(u^2+1)}=dfrac Au+dfrac{Bu+C}{u^2+1}$



                  $implies1=A(u^2+1)+u(Bu+C)=u^2(A+B)+Cu+A$



                  Comparing the coefficients of $u,u^2$ and the constants



                  $implies C=0,A=1,A+B=0iff B=-A=?$



                  Can you take it from here?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 16:16









                  lab bhattacharjeelab bhattacharjee

                  228k15158279




                  228k15158279






























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