integral 5x cos3x cosx dx

Integration by Trigonometry Identity and by Parts.

Remember, cos α cos β = 1/2 [cos (α + β) + cos (α - β)]

∫ 5x cos 3x cos x dx
= 5 ∫ x cos 3x cos x dx
= 5 ∫ x {1/2 [cos (3x + x) + cos (3x - x)]} dx
= 5/2 ∫ (x cos 4x + x cos 2x) dx

For ∫ x cos 4x dx
u = x → du = dx
dv = cos 4x = v = 1/4 sin 4x
∫ u dv = uv - ∫ v du
∫ x cos 4x dx = 1/4 x sin 4x - ∫ 1/4 sin 4x dx
                     = 1/4 x sin 4x - 1/4 (-1/4 cos 4x)
                     = 1/4 x sin 4x + 1/16 cos 4x

For ∫ x cos 2x
u = x → du = dx
dv = cos 2x → v = 1/2 sin 2x
∫ u dv = uv - ∫ v du
∫ x cos 2x = 1/2 x sin 2x - ∫ 1/2 sin 2x dx
                = 1/2 x sin 2x - 1/2 (-1/2 cos 2x) dx
                = 1/2 x sin 2x + 1/4 cos 2x

Plug in solved integrals!
= 5/2 ∫ (x cos 4x + x cos 2x) dx
= 5/2 (1/4 x sin 4x + 1/16 cos 4x + 1/2 x sin 2x + 1/4 cos 2x) + C
= 5/32 (4x sin 4x + cos 4x + 8x sin 2x + 4 cos 2x) + C
= 5/32 [4 cos 2x + cos 4x + 4x (sin 4x + 2 sin 2x)] + C
= 5/32 [4 cos 2x + cos 4x + 4x (2 sin 2x cos 2x + 4 sin x cos x)] + C
= 5/32 {4 cos 2x + cos 4x + 4x [4 sin x cos x (2 cos² x - 1) + 4 sin x cos x]}+ C
= 5/32 (4 cos 2x + cos 4x + 32 x sin x cos³ x) + C