How to solve $int sin^3(x) cos^2(x) dx$ with integration by parts?












0














Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried



$$int sin^3(x) cos^2(x)dx$$



Let: $u'=cos (x), u = sin (x), v = sin (x) , v'=cos (x)$



$$int v^3 u'^2$$



And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents






calculus real-analysis integration trigonometry







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  • 2




    dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
    – mathreadler
    Jan 31 '18 at 20:16


















0














Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried



$$int sin^3(x) cos^2(x)dx$$



Let: $u'=cos (x), u = sin (x), v = sin (x) , v'=cos (x)$



$$int v^3 u'^2$$



And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents






calculus real-analysis integration trigonometry







share|cite|improve this question




















  • 2




    dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
    – mathreadler
    Jan 31 '18 at 20:16
















0












0








0







Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried



$$int sin^3(x) cos^2(x)dx$$



Let: $u'=cos (x), u = sin (x), v = sin (x) , v'=cos (x)$



$$int v^3 u'^2$$



And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents






calculus real-analysis integration trigonometry







share|cite|improve this question















Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried



$$int sin^3(x) cos^2(x)dx$$



Let: $u'=cos (x), u = sin (x), v = sin (x) , v'=cos (x)$



$$int v^3 u'^2$$



And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents






calculus real-analysis integration trigonometry




calculus real-analysis integration trigonometry






share|cite|improve this question















share|cite|improve this question













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edited Jan 31 '18 at 20:25









Michael Rozenberg

97.4k1589188




97.4k1589188










asked Jan 31 '18 at 20:11









TreyTrey

304113




304113








  • 2




    dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
    – mathreadler
    Jan 31 '18 at 20:16
















  • 2




    dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
    – mathreadler
    Jan 31 '18 at 20:16










2




2




dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
– mathreadler
Jan 31 '18 at 20:16






dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense
– mathreadler
Jan 31 '18 at 20:16












5 Answers
5






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1














split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$






share|cite|improve this answer





























    2















    $$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$







    share|cite|improve this answer





























      1














      $$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
      $$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
      $$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$






      share|cite|improve this answer





























        1














        Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.



        $int sin^3x cos^2 x dx$



        pick something easy to integrate to be $v'$



        $v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.



        lets go with the first one.



        $u = sin^2 x, dv = sin xcos^2 x dx\
        du = 2sin xcos x dx, v = -frac 13 cos^3x$



        $-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
        -frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
        frac 15cos^5 x - frac 13 cos^3 x$






        share|cite|improve this answer





























          0














          Integration by parts is unnecessary. Do this
          $$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
          Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.






          share|cite|improve this answer





















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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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            1














            split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
            furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
            and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$






            share|cite|improve this answer


























              1














              split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
              furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
              and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$






              share|cite|improve this answer
























                1












                1








                1






                split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
                furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
                and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$






                share|cite|improve this answer












                split your integral and write $$sin(x)^3(1-sin(x)^2)=sin(x)^3-sin(x)^5$$
                furthere use that $$sin(x)^3=frac{1}{4} (3 sin (x)-sin (3 x))$$
                and $$sin(x)^5=frac{1}{16} (10 sin (x)-5 sin (3 x)+sin (5 x))$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 31 '18 at 20:14









                Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                73.5k42865




                73.5k42865























                    2















                    $$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$







                    share|cite|improve this answer


























                      2















                      $$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$







                      share|cite|improve this answer
























                        2












                        2








                        2







                        $$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$







                        share|cite|improve this answer













                        $$int sin ^{ 3 }{ x } cos ^{ 2 }{ x } dx=-int { sin ^{ 2 }{ x } cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =-int { left( 1-cos ^{ 2 }{ x } right) cos ^{ 2 }{ x } dleft( cos { x } right) } =int { cos ^{ 4 }{ x } -cos ^{ 2 }{ x } dleft( cos { x } right) } =\ =frac { cos ^{ 5 }{ x } }{ 5 } -frac { cos ^{ 3 }{ x } }{ 3 } +C$$








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                        answered Jan 31 '18 at 20:15









                        haqnaturalhaqnatural

                        20.6k72457




                        20.6k72457























                            1














                            $$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
                            $$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
                            $$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$






                            share|cite|improve this answer


























                              1














                              $$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
                              $$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
                              $$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$






                              share|cite|improve this answer
























                                1












                                1








                                1






                                $$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
                                $$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
                                $$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$






                                share|cite|improve this answer












                                $$intsin^3xcos^2xdx=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}intsin^4x(-sin{x})dx=$$
                                $$=frac{sin^4x}{4}cdotcos{x}-frac{1}{4}int(1-cos^2x)^2d(cos{x})=$$
                                $$=frac{sin^4xcos{x}}{4}-frac{cos{x}}{4}+frac{cos^3x}{6}-frac{cos^5x}{20}+C.$$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 31 '18 at 20:17









                                Michael RozenbergMichael Rozenberg

                                97.4k1589188




                                97.4k1589188























                                    1














                                    Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.



                                    $int sin^3x cos^2 x dx$



                                    pick something easy to integrate to be $v'$



                                    $v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.



                                    lets go with the first one.



                                    $u = sin^2 x, dv = sin xcos^2 x dx\
                                    du = 2sin xcos x dx, v = -frac 13 cos^3x$



                                    $-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
                                    -frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
                                    frac 15cos^5 x - frac 13 cos^3 x$






                                    share|cite|improve this answer


























                                      1














                                      Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.



                                      $int sin^3x cos^2 x dx$



                                      pick something easy to integrate to be $v'$



                                      $v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.



                                      lets go with the first one.



                                      $u = sin^2 x, dv = sin xcos^2 x dx\
                                      du = 2sin xcos x dx, v = -frac 13 cos^3x$



                                      $-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
                                      -frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
                                      frac 15cos^5 x - frac 13 cos^3 x$






                                      share|cite|improve this answer
























                                        1












                                        1








                                        1






                                        Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.



                                        $int sin^3x cos^2 x dx$



                                        pick something easy to integrate to be $v'$



                                        $v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.



                                        lets go with the first one.



                                        $u = sin^2 x, dv = sin xcos^2 x dx\
                                        du = 2sin xcos x dx, v = -frac 13 cos^3x$



                                        $-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
                                        -frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
                                        frac 15cos^5 x - frac 13 cos^3 x$






                                        share|cite|improve this answer












                                        Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.



                                        $int sin^3x cos^2 x dx$



                                        pick something easy to integrate to be $v'$



                                        $v' = sin xcos^2 x$ or $v' = sin^3xcos x$ would be my suggestions.



                                        lets go with the first one.



                                        $u = sin^2 x, dv = sin xcos^2 x dx\
                                        du = 2sin xcos x dx, v = -frac 13 cos^3x$



                                        $-frac 13 cos^3 xsin^2 x + frac 23 int cos^4 xsin x dx\
                                        -frac 13 cos^3 xsin^2 x - frac 2{15} cos^5 x\
                                        frac 15cos^5 x - frac 13 cos^3 x$







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Dec 12 '18 at 22:09









                                        Doug MDoug M

                                        44.2k31854




                                        44.2k31854























                                            0














                                            Integration by parts is unnecessary. Do this
                                            $$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
                                            Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.






                                            share|cite|improve this answer


























                                              0














                                              Integration by parts is unnecessary. Do this
                                              $$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
                                              Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.






                                              share|cite|improve this answer
























                                                0












                                                0








                                                0






                                                Integration by parts is unnecessary. Do this
                                                $$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
                                                Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.






                                                share|cite|improve this answer












                                                Integration by parts is unnecessary. Do this
                                                $$int sin^3(x)cos^2(x),dx = int (1-cos^2(x))cos^2(x)sin(x),dx $$
                                                Let $u = cos(x), du = -sin(x),dx$ and this resolves very simply.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Dec 12 '18 at 21:52









                                                ncmathsadistncmathsadist

                                                42.4k259102




                                                42.4k259102






























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