kak tolong dikerjakan soal integral

∫ sin ax cos bx dx
= ∫ 1/2 (sin (ax + bx) + sin (ax - bx)) dx
= ∫ [1/2 sin (a + b)x + 1/2 sin (a - b)x] dx
= 1/2 . -1/(a + b) cos (a + b)x + 1/2 . -1/(a - b) cos (a - b)x + C
= -1/(2a + 2b) cos (a + b)x - 1/(2a - 2b) cos (a - b)x + C

= -1/(2(a + b)) cos (a + b)x - 1/(2(a - b)) cos (a - b)x + C

= -1/(2(a + b)) cos (ax + bx) - 1/(2(a - b)) cos (ax - bx)