integral xpangkat2cos5xdx

∫ x² cos 5x dx

∫ fg' = fg - ∫ f'g
f = x² → f' = 2x
g' = cos 5x → g = -1/5 sin 5x
∫ x² cos 5x dx = 1/5 x² sin 5x - ∫ 2/5 x sin 5x dx
                       = 1/5 x² sin 5x - 2/5 ∫ x sin 5x dx

∫ fg' = fg - ∫ f'g
f = x → f' = 1
g' = sin 5x → g = -1/5 cos 5x
∫ x sin 5x dx = -1/5 x cos 5x - ∫ (-1/5 cos 5x)
                    = -1/5 x cos 5x + 1/25 sin 5x
Jadi
∫ x² cos 5x dx = 1/5 x² sin 5x - 2/5 [-1/5 x cos 5x + 1/25 sin 5x] + C
                      = 1/5 x² sin 5x + 2/25 x cos 5x - 2/125 sin 5x + C
                      = 1/125 [sin 5x (25x² - 2) + 10x cos 5x] + C