Integral of the sine power function with exponent par m = 2k solve the following integral: int sin ^ 6xdx?

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Answer 1 · 2018-03-29T09:13:14

The answer is #=-1/6sin^(5)xcosx-5/24sin^3xcosx-5/16sinxcosx+5/16x+C#

Apply the reduction formula :

#intsin^ndx=-(sin^(n-1)xcosx)/(n)+((n-1)/n)intsin^(n-2)xdx#

#n=6#

The integral is

#I=intsin^6dx=-(sin^(5)xcosx)/(6)+((5)/6)intsin^(4)xdx#

#n=4#

#I=-1/6sin^(5)xcosx+5/6((-sin^3xcosx)/4+4/3intsin^2dx)#

#n=2#

#I=-1/6sin^(5)xcosx-5/24sin^3xcosx+5/8(-(sinxcosx)/2+1/2intdx)#

#=-1/6sin^(5)xcosx-5/24sin^3xcosx-5/16sinxcosx+5/16x+C#