How do you find #intsin^2 (2x) cos^3 (2x) dx#?

Question

7537 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-13T09:57:14

#(sin2x)^3/6-(sin2x)^5/10+C#

#int (sin2x)^2*(cos2x)^3*dx#

=#int (sin2x)^2*(cos2x)^2*cos2x*dx#

=#int (sin2x)^2*(1-(sin2x)^2)*cos2x*dx#

=#1/2int (sin2x)^2*(1-(sin2x)^2)*2cos2x*dx#

After using #y=sin2x# and #dy=2cos2y*dy# transforms, this integral became

#1/2int y^2*(1-y^2)*dy#

=#1/2int y^2*dy-1/2int y^4*dy#

=#y^3/6-y^5/10+C#

=#(sin2x)^3/6-(sin2x)^5/10+C#