0
= 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
Wiki User
∙ 10y ago
Wiki User
∙ 10y agoInt(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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