Buktikan bahwa [tex]\bold{ \int\limits {u.cos^2(u).sin(u)} \, du}= [/tex][tex] \dfrac{1}{36}(9sin(u)+sin(3u)-9u.cos(u)-3u.cos(3u)) +C[/tex]

Saya gak pakai latex, ya

cos²(u) = 1 - sin²(u)

∫u.cos²(u).sin(u) du = ∫u.(1 - sin²(u)).sin(u) du
= ∫u(sin(u) - sin³(u)) du
= ∫u.sin(u) du - ∫u.sin³(u) du

bagi jadi dua pengerjaan.

yang pertama,
∫u.sin(u) du

pakai integral parsial.
misal, u = u
du = du

dv = sin(u) du
v = -cos(u)

uv - ∫vdu = u.(-cos(u)) - ∫(-cos(u)) du
= -u.cos(u) + ∫cos(u) du
= -u.cos(u) + sin(u)

untuk integral kedua

∫u.sin³(u) du

misal,
u = u
du = du
dv = sin³u du
v = (1/12)(cos(3u) - 9cos(u)) (sin³u nya tinggal diubah pakai identitas trigonometri, terus hasilnya diubah bentuknya jadi kayak gini).

uv - ∫vdu = u(1/12)(cos(3u) - 9cos(u)) - ∫(1/12)(cos(3u) - 9cos(u)) du
= (u/12)(cos(3u) - 9cos(u)) - (1/36)sin(3u) + (9/12)sin(u)
= (u/12)cos(3u) - (9u/12)cos(u) - (1/36)sin(3u) + (9/12)sin(u)

jadi,

∫u.sin(u) du - ∫u.sin³(u) du
=
-u.cos(u) + sin(u) - ((u/12)cos(3u) - (9u/12)cos(u) - (1/36)sin(3u) + (9/12)sin(u)) + C
= (-u + 9u/12)cos(u) + (1 - 9/12)sin(u) - (u/12)cos(3u) + (1/36)sin(3u) + C
= (-3u/12)cos(u) + (3/12)sin(u) - (u/12)cos(3u) + (1/36)sin(3u) + C
= (-9u/36)cos(u) + (9/36)sin(u) - (3u/12)cos(3u) + (1/36)sin(3u) + C
= (1/36)(9sin(u) + sin(3u) - 9u.cos(u) - 3u.cos(3u)) + C