1. a) Determine \alpha e \beta de modo que
\begin{array}{r} \operatorname{sen} 3 x \operatorname{sen} 2 x=-\frac{1}{2}(\cos \alpha x-\cos \beta x) \\\ \left(\text { Sugestão }: \operatorname{sen} a \operatorname{sen} b=\frac{1}{2}[\cos (a-b)-\cos (a+b)] .\right. \end{array}ã
b) Calcule \int \operatorname{sen} 3 x \operatorname{sen} 2 x d x.

Passo 1

O enunciado pede duas coisas:
(a) Expressar \operatorname{sen}3x\,\operatorname{sen}2x na forma
-\tfrac12\bigl(\cos(\alpha x)-\cos(\beta x)\bigr)
usando a identidade de produto para soma.
(b) Calcular \displaystyle\int \operatorname{sen}3x\,\operatorname{sen}2x\,dx a partir dessa decomposição.

Passo 2

Aplicamos a identidade sugerida, que afirma
\operatorname{sen}a\,\operatorname{sen}b =\tfrac12\bigl[\cos(a-b)-\cos(a+b)\bigr].
Aqui a=3x e b=2x, logo
\operatorname{sen}3x\,\operatorname{sen}2x =\tfrac12\bigl[\cos(3x-2x)-\cos(3x+2x)\bigr] =\tfrac12\bigl[\cos x-\cos5x\bigr].

Passo 3

Queremos escrever
\operatorname{sen}3x\,\operatorname{sen}2x =-\tfrac12\bigl(\cos(\alpha x)-\cos(\beta x)\bigr).
Como
\tfrac12\bigl[\cos x-\cos5x\bigr] =-\tfrac12\bigl[\cos5x-\cos x\bigr],
vemos que
-\tfrac12\bigl(\cos(\alpha x)-\cos(\beta x)\bigr) =-\tfrac12\bigl[\cos5x-\cos x\bigr]
implica \alpha=5 e \beta=1.

Passo 4

Para a integral, usamos a forma mais simples já obtida:
\int \operatorname{sen}3x\,\operatorname{sen}2x\,dx =\int \tfrac12\bigl(\cos x-\cos5x\bigr)\,dx =\tfrac12\int\cos x\,dx-\tfrac12\int\cos5x\,dx.
Calculando cada termo:
\int\cos x\,dx=\sin x,\qquad \int\cos5x\,dx=\tfrac{1}{5}\sin5x.
Logo,
\int \operatorname{sen}3x\,\operatorname{sen}2x\,dx =\tfrac12\sin x-\tfrac12\cdot\tfrac{1}{5}\sin5x + C =\tfrac12\sin x-\tfrac{1}{10}\sin5x + C.

Resposta

(a) \alpha=5, \beta=1.
(b) \displaystyle\int \operatorname{sen}3x\,\operatorname{sen}2x\,dx =\tfrac12\sin x-\tfrac{1}{10}\sin5x + C.

Exercícios relacionados