Thread: integration 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

Originally Posted by HyrdaD

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

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

That means use the substitution t = cos x.

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

Note,
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.