Use integration by parts to deduce the following formulas.

A) int_ sin^3[x] dx = (-3/4)cos[x] + (1/12)cos[3x]

B) int_ sin^4[x] dx = (3/8)x -(1/4)sin[2x] + (1/32)sin[4x]

Thanks so much

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- May 20th 2007, 11:35 PMHyrdaDintegration by parts
Use integration by parts to deduce the following formulas.

A) int_ sin^3[x] dx = (-3/4)cos[x] + (1/12)cos[3x]

B) int_ sin^4[x] dx = (3/8)x -(1/4)sin[2x] + (1/32)sin[4x]

Thanks so much - May 21st 2007, 05:56 AMThePerfectHacker
Note,

sin^3 x = sin^2 x * sin x = (1 - cos^2 x)*sin x

That means use the substitution t = cos x.

Quote:

B) int_ sin^4[x] dx = (3/8)x -(1/4)sin[2x] + (1/32)sin[4x]

sin^4 x = (sin^2 x)^2 = (1/2 - 1/2*cos 2x)^2

Expand,

1/4 - 1/2*cos 2x + 1/4*cos^2 (2x)

You can use another half-angle formula on cosine.