trigonometric integration problem












1














Hello guys can someone please help me find the answer :




$$int sin x tan x~dx$$







indefinite-integrals







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    1














    Hello guys can someone please help me find the answer :




    $$int sin x tan x~dx$$







    indefinite-integrals







    share|cite|improve this question



























      1












      1








      1







      Hello guys can someone please help me find the answer :




      $$int sin x tan x~dx$$







      indefinite-integrals







      share|cite|improve this question















      Hello guys can someone please help me find the answer :




      $$int sin x tan x~dx$$







      indefinite-integrals




      indefinite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 41 mins ago









      mrtaurho

      3,1051930




      3,1051930










      asked 42 mins ago









      RS 2838484

      61




      61






















          2 Answers
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          7














          Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$






          share|cite|improve this answer































            1














            Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
            $$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
            the integral of a rational function. Thus it comes down to a decomposition into partial fractions.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

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              7














              Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$






              share|cite|improve this answer




























                7














                Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$






                share|cite|improve this answer


























                  7












                  7








                  7






                  Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$






                  share|cite|improve this answer














                  Note that $$sin(x)tan(x)=frac{sin^2(x)}{cos(x)}=frac{1-cos^2(x)}{cos(x)}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 22 mins ago









                  Bernard

                  117k637111




                  117k637111










                  answered 38 mins ago









                  Dr. Sonnhard Graubner

                  72.6k32865




                  72.6k32865























                      1














                      Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
                      $$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
                      the integral of a rational function. Thus it comes down to a decomposition into partial fractions.






                      share|cite|improve this answer


























                        1














                        Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
                        $$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
                        the integral of a rational function. Thus it comes down to a decomposition into partial fractions.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
                          $$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
                          the integral of a rational function. Thus it comes down to a decomposition into partial fractions.






                          share|cite|improve this answer












                          Bioche's rules say you should make the substitution $;u=sin x$, $mathrm d u=cos x,mathrm dx$, so that
                          $$int sin x tan x,mathrm dx=intfrac{sin^2x}{cos x},frac{mathrm du}{cos x}=intfrac{u^2}{1-u^2},mathrm du,$$
                          the integral of a rational function. Thus it comes down to a decomposition into partial fractions.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 13 mins ago









                          Bernard

                          117k637111




                          117k637111






























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