Question

Tentukan integral berikut dengan teknik pengintegralan bagian demi bagian :
a. integral x² e^(-2x) dx
b. integral x sin x dx​

Answer 1

Jawaban:

a. ∫x² e^(-2x) dx

= ∫x² (-2) e^(-2x) dx

= -2 ∫x² e^(-2x) dx

= -2 [x² (-1/2) e^(-2x) + (1/2) ∫2x e^(-2x) dx]

= -x² (-1/2) e^(-2x) - (1/2) ∫2 e^(-2x) dx

= -x² (-1/2) e^(-2x) - (1/2) [2 (-1/2) e^(-2x) + (1/2) ∫e^(-2x) dx]

= -x² (-1/2) e^(-2x) - 2 (-1/2) e^(-2x) - (1/4) ∫e^(-2x) dx

= -x² (-1/2) e^(-2x) - 2 (-1/2) e^(-2x) - (1/4) [e^(-2x) (-1/2) + (1/2) ∫1 dx]

= -x² (-1/2) e^(-2x) - 2 (-1/2) e^(-2x) - (1/4) e^(-2x) (-1/2) + (1/4) x

= -x² (-1/2) e^(-2x) - e^(-2x) (-1/2) + (1/4) x + C

b. ∫x sin x dx

= ∫x (-cos x) dx

= - ∫x cos x dx

= - [x (-1/2) sin x + (1/2) ∫2 sin x dx]

= -x (-1/2) sin x - (1/2) ∫2 (-cos x) dx

= -x (-1/2) sin x - (1/2) [2 (-1/2) sin x + (1/2) ∫1 (-cos x) dx]

= -x (-1/2) sin x - sin x (-1) - (1/4) ∫(-cos x) dx

= -x (-1/2) sin x - sin x (-1) - (1/4) [(-cos x) (-1/2) + (1/2) ∫1 dx]

= -x (-1/2) sin x - sin x (-1) - (-cos x) (-1/2) + (1/4) x + C

= x (-1/2) sin x + sin x + (1/4) x + C