integral x sin^3 x dx

Halo, kakak bantu jawab ya. Jawaban untuk soal di atas adalah ∫ x sin³x dx = ∫ x sin³x dx = - x cos x + (1/3)xcos³x + (2/3) sin x + (1/9) sin³x + C Kita ingat dulu bentuk integral berikut : ∫ u dv = u.v - ∫ v du ∫ sin x dx = -cos x + C ∫ cos x dx = sinx + C ∫ sinⁿx cos x dx = (1/(n+1)) sin⁽ⁿ⁺¹⁾x + C ∫ cosⁿx sin x dx = - (1/(n+1)) cos⁽ⁿ⁺¹⁾x + C Misalkan : u = x --> du /dx = 1 --> du = dx dv = sin³x dx --> v = ∫ sin³x dx = ∫sin x (sin²x) dx = ∫ sin x (1-cos²x) dx = ∫ sin x - cos²x sin x dx = - cos x + (1/3)cos³x maka : ∫ x sin 3x dx = ∫ u dv = uv - ∫ v du = x(- cos x + (1/3)cos³x ) - ∫(- cos x + (1/3)cos³x) dx = -x cos x + (1/3)xcos³x - ∫(- cos x + (1/3)cosx(cos²x)) dx = -x cos x + (1/3)xcos³x - ∫(- cos x + (1/3)cosx(1-sin²x)) dx = -x cos x + (1/3)xcos³x - ∫(- cos x + (1/3)cosx -(1/3)sin²x cos x)) dx = -x cos x + (1/3)xcos³x - (-sinx + (1/3)sin x - (1/9)sin³x ) + C = -x cos x + (1/3)xcos³x - (-sinx + (1/3)sin x - (1/9)sin³x ) + C = -x cos x + (1/3)xcos³x + sin x - (1/3) sin x + (1/9) sin³x + C = -x cos x + (1/3)xcos³x + (3/3) sin x - (1/3) sin x + (1/9) sin³x + C = -x cos x + (1/3)xcos³x + (2/3) sin x + (1/9) sin³x + C Jadi nilai dari ∫ x sin³x dx = -x cos x + (1/3)xcos³x + (2/3) sin x + (1/9) sin³x + C Terimakasih sudah bertanya