Question #c3e4f

Question

1284 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-05T19:38:41

#int (sin6x*dx)/sinx=2/5*sin5x+2/3*sin3x+2sinx+C#

I decomposed #sin6x#,

#sin6x#

#=sin6x-sin2x+sin2x#

#=2cos4x*sin2x+sin2x#

#=sin2x*(2cos4x+1)#

Hence,

#(sin6x*dx)/sinx#

=#[sin2x*(2cos4x+1)*dx]/sinx#

=#[2sinx*cosx*(2cos4x+1)*dx]/sinx#

=#2cosx*(2cos4x+1)*dx#

=#int 4cos4x*cosx*dx#+#int 2cosx*dx#

=#int 2cos5x*dx#+#int 2cos3x*dx#+#int 2cosx*dx#

=#2/5*sin5x+2/3*sin3x+2sinx+C#