Resolva as integrais:
\int \sin (3 x) \cos \left(\frac{1}{2} x\right) d x

Passo 1

O enunciado pede para calcular a integral
\int \sin(3x)\,\cos\!\bigl(\tfrac12x\bigr)\,dx.
O conteúdo da matéria é integração de produtos de funções trigonométricas, e vamos usar a identidade produto-para-soma para simplificar antes de integrar.

Passo 2

Aplicamos a identidade
\sin A\cos B = \tfrac12\bigl[\sin(A+B)+\sin(A-B)\bigr]
com A=3x e B=\tfrac12x. Então
\sin(3x)\cos\!\bigl(\tfrac12x\bigr) = \tfrac12\Bigl[\sin\!\bigl(3x+\tfrac12x\bigr)+\sin\!\bigl(3x-\tfrac12x\bigr)\Bigr] = \tfrac12\bigl[\sin\!\bigl(\tfrac72x\bigr)+\sin\!\bigl(\tfrac52x\bigr)\bigr].

Passo 3

Substituindo na integral, obtemos
\int \sin(3x)\cos\!\bigl(\tfrac12x\bigr)\,dx = \tfrac12\int \bigl[\sin\!\bigl(\tfrac72x\bigr)+\sin\!\bigl(\tfrac52x\bigr)\bigr]dx.

Passo 4

Integramos cada termo usando \int \sin(kx)\,dx = -\tfrac1k\cos(kx):
\int \sin\!\bigl(\tfrac72x\bigr)\,dx = -\tfrac{2}{7}\cos\!\bigl(\tfrac72x\bigr), \qquad \int \sin\!\bigl(\tfrac52x\bigr)\,dx = -\tfrac{2}{5}\cos\!\bigl(\tfrac52x\bigr).

Passo 5

Multiplicando por \tfrac12 e somando as constantes, vem
\int \sin(3x)\cos\!\bigl(\tfrac12x\bigr)\,dx = \tfrac12\Bigl[-\tfrac{2}{7}\cos\!\bigl(\tfrac72x\bigr)-\tfrac{2}{5}\cos\!\bigl(\tfrac52x\bigr)\Bigr]+C = -\tfrac{1}{7}\cos\!\bigl(\tfrac72x\bigr)-\tfrac{1}{5}\cos\!\bigl(\tfrac52x\bigr)+C.

Resposta

\int \sin(3x)\cos\!\bigl(\tfrac12x\bigr)\,dx = -\tfrac{1}{7}\cos\!\bigl(\tfrac72x\bigr)-\tfrac{1}{5}\cos\!\bigl(\tfrac52x\bigr)+C.

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