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= cos(x)-(cos3(x))/3

* * * * *

Right numbers, wrong sign!

Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx

= Int(sinx)dx + Int[-cos2x*sinx]dx

Int(sinx)dx = -cosx . . . . . (I)

Int[-cos2x*sinx]dx

Let u = cosx, the du = -sinxdx

so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II)

So Int(sin3x)dx = 1/3*cos3x - cosx + C

Alternatively,

using the multiple angle identities, you can show that

sin3x = 1/4*[3sinx - sin3x]

which gives

Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C

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11y ago

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