How do you find the integral of # sin^3[x]dx#?

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Answer 1 · 2016-03-19T18:44:23

#intsin^3(x)dx = 1/3cos^3(x)-cos(x)+C#

#intsin^3(x)dx = intsin(x)(1-cos^2(x))dx#

#=intsin(x)dx - intsin(x)cos^2(x)dx#

For the first integral:

#intsin(x)dx = -cos(x)+C#

For the second integral, using substitution:

Let #u = cos(x) => du = -sin(x)dx#
Then

#-intsin(x)cos^2(x)dx = intu^2du#

#=u^3/3+C#

#=1/3cos^3(x)+C#

Putting it all together, we get our final result:

#intsin^3(x)dx = intsin(x)dx-intsin(x)cos^2(x)dx#

#=-cos(x)+1/3cos^3(x)+C#