Question #c3e4f

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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,

$(sin6x*dx)/sinx$

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

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

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

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

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

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