Calcule a integral.
\int e^{x} \sin \beta t d t

Passo 1

O enunciado pede para calcular
\int e^{x}\sin(\beta x)\,dx
O conteúdo principal é integração por partes, e vamos precisar aplicá-la duas vezes para voltar a obter a integral original e, então, resolver a equação resultante.

Passo 2

Defina
I = \int e^{x}\sin(\beta x)\,dx.
Aplique integração por partes escolhendo
u = e^{x},\quad dv = \sin(\beta x)\,dx.
Então
du = e^{x}\,dx, \quad v = \int \sin(\beta x)\,dx = -\frac{\cos(\beta x)}{\beta}.
Logo,
I = u\,v - \int v\,du = -\frac{e^{x}\cos(\beta x)}{\beta} + \frac{1}{\beta}\int e^{x}\cos(\beta x)\,dx.
Chame
J = \int e^{x}\cos(\beta x)\,dx,
de modo que
I = -\frac{e^{x}\cos(\beta x)}{\beta} + \frac{1}{\beta}\,J.

Passo 3

Calcule agora
J = \int e^{x}\cos(\beta x)\,dx
por partes novamente, tomando
u = e^{x},\quad dv = \cos(\beta x)\,dx.
Temos
du = e^{x}\,dx, \quad v = \int \cos(\beta x)\,dx = \frac{\sin(\beta x)}{\beta}.
Então
J = u\,v - \int v\,du = \frac{e^{x}\sin(\beta x)}{\beta} - \frac{1}{\beta}\int e^{x}\sin(\beta x)\,dx = \frac{e^{x}\sin(\beta x)}{\beta} - \frac{I}{\beta}.

Passo 4

Substitua a expressão de J em
I = -\frac{e^{x}\cos(\beta x)}{\beta} + \frac{1}{\beta}\,J.
Isso dá
I = -\frac{e^{x}\cos(\beta x)}{\beta} + \frac{1}{\beta}\Bigl(\frac{e^{x}\sin(\beta x)}{\beta} - \frac{I}{\beta}\Bigr) = -\frac{e^{x}\cos(\beta x)}{\beta} + \frac{e^{x}\sin(\beta x)}{\beta^{2}} - \frac{I}{\beta^{2}}.
Rearranjando para isolar I:
I + \frac{I}{\beta^{2}} = e^{x}\Bigl(\frac{\sin(\beta x)}{\beta^{2}} - \frac{\cos(\beta x)}{\beta}\Bigr).
Como
1 + \frac{1}{\beta^{2}} = \frac{\beta^{2}+1}{\beta^{2}},
temos
I = e^{x}\Bigl(\frac{\sin(\beta x)}{\beta^{2}} - \frac{\cos(\beta x)}{\beta}\Bigr)\cdot\frac{\beta^{2}}{\beta^{2}+1} = \frac{e^{x}\bigl(\sin(\beta x) - \beta\cos(\beta x)\bigr)}{\beta^{2}+1}.

Resposta

\int e^{x}\sin(\beta x)\,dx = \frac{e^{x}\bigl(\sin(\beta x) - \beta\cos(\beta x)\bigr)}{\beta^{2}+1} + C.

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