Integrais trigonometricas: \int \sin ^{m}(p x) \cos ^{n}(q x) d x
\sin p x \cos q x=\frac{1}{2} \sin (p+q) x+\frac{1}{2} \sin (p-q) x
\sin p x \sin q x=-\frac{1}{2} \cos (p+q) x+\frac{1}{2} \cos (p-q) x
\cos p x \cos q x=\frac{1}{2} \cos (p+q) x+\frac{1}{2} \cos (p-q) x
4. \int_{0}^{\pi} \cos x \sin 2 x d x

Passo 1

O enunciado pede calcular a integral definida
\int_{0}^{\pi} \cos x \,\sin 2x \,dx
usando identidades de integrais trigonométricas, em particular
\sin p x \cos q x = \tfrac12\bigl[\sin\bigl(p+q\bigr)x + \sin\bigl(p-q\bigr)x\bigr].

Passo 2

Aplicando a identidade com p=2 e q=1, obtemos
\sin 2x \cos x = \tfrac12\bigl[\sin(3x) + \sin(x)\bigr].
Logo
\int_{0}^{\pi} \cos x \,\sin 2x \,dx = \tfrac12\int_{0}^{\pi} \bigl[\sin(3x)+\sin(x)\bigr]dx.

Passo 3

Vamos calcular a integral indefinida:
\int \bigl[\sin(3x)+\sin(x)\bigr]dx = -\tfrac{\cos(3x)}{3} - \cos(x) + C.
Multiplicando por \tfrac12, a primitiva é
F(x) = -\tfrac{\cos(3x)}{6} - \tfrac{\cos(x)}{2}.

Passo 4

Agora avaliamos de 0 a \pi:
F(\pi) = -\tfrac{\cos(3\pi)}{6} - \tfrac{\cos(\pi)}{2} = -\tfrac{(-1)}{6} - \tfrac{(-1)}{2} = \tfrac{1}{6} + \tfrac{1}{2} = \tfrac{2}{3},
F(0) = -\tfrac{\cos(0)}{6} - \tfrac{\cos(0)}{2} = -\tfrac{1}{6} - \tfrac{1}{2} = -\tfrac{2}{3}.
Portanto
\int_{0}^{\pi} \cos x \,\sin 2x \,dx = F(\pi) - F(0) = \tfrac{2}{3} - \bigl(-\tfrac{2}{3}\bigr) = \tfrac{4}{3}.

Resposta

\displaystyle \frac{4}{3}

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