Avalie a integral.
integral t^2 sin (beta t) d t

Passo 1

O enunciado pede para calcular a integral integral t^2sin(beta t),dt. O conteúdo principal é integração por partes, que usa a fórmula integral u,dv = u v - integral v,du.

Passo 2

Escolhemos u = t^2 e dv = sin(beta t),dt. Então
$$du = 2t,dt,
$$
$$v = \int \sin(\beta t),dt = -\frac{1}{\beta}\cos(\beta t).
du = 2t,dt, çã\int t^2\sin(\beta t),dt = u v - \int v,du = -\frac{t^2}{\beta}\cos(\beta t) + \frac{2}{\beta}\int t\cos(\beta t),dt.
$$

Passo 3

Para calcular integral tcos(beta t),dt, aplicamos IBP de novo: escolhemos u1 = t e dv1 = cos(beta t),dt. Então
$$du_1 = dt,
$$
$$v_1 = \int \cos(\beta t),dt = \frac{1}{\beta}\sin(\beta t).
du1 = dt, \int t\cos(\beta t),dt = u_1 v_1 - \int v_1,du_1 = \frac{t}{\beta}\sin(\beta t) - \int \frac{1}{\beta}\sin(\beta t),dt = \frac{t}{\beta}\sin(\beta t) + \frac{1}{\beta^2}\cos(\beta t).
$$

Passo 4

Substituindo de volta na expressão original:
$$\int t^2\sin(\beta t),dt = -\frac{t^2}{\beta}\cos(\beta t) + \frac{2}{\beta}\Bigl(\frac{t}{\beta}\sin(\beta t) + \frac{1}{\beta^2}\cos(\beta t)\Bigr).
integral t^2sin(beta t),dt = -t^2betacos(beta t) + 2betal(tbetasin(beta t) + 1beta^2cos(beta t)r) \int t^2\sin(\beta t),dt = -\frac{t^2}{\beta}\cos(\beta t) + \frac{2t}{\beta^2}\sin(\beta t) + \frac{2}{\beta^3}\cos(\beta t)$$

Passo 5

Resposta

-t^2betacos(beta t) + 2tbeta^2sin(beta t) + 2beta^3cos(beta t) + C

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