Questão. ( \mathbf{4 , 0 ~ p t s )}
Calcule as integrais.
(b) \int_{0}^{\frac{\pi}{4}} \operatorname{sen}^{3}(2 x) \cos ^{2}(2 x) d x

Passo 1

O enunciado pede para calcular a integral
\int_{0}^{\frac{\pi}{4}}\sin^{3}(2x)\,\cos^{2}(2x)\,dx.
O conteúdo principal é usar identidades trigonométricas para reescrever o integrando e depois aplicar substituição simples.

Passo 2

Repare que
\sin^{3}(2x)=\sin^{2}(2x)\,\sin(2x)=(1-\cos^{2}(2x))\,\sin(2x).
Então o integrando vira
\sin^{3}(2x)\,\cos^{2}(2x) =(1-\cos^{2}(2x))\,\sin(2x)\,\cos^{2}(2x).

Passo 3

Façamos a substituição
u=\cos(2x), \quad du=-2\,\sin(2x)\,dx \;\Longrightarrow\; \sin(2x)\,dx=-\tfrac{1}{2}\,du.
Logo
\int_{0}^{\frac{\pi}{4}}(1-\cos^{2}(2x))\,\sin(2x)\,\cos^{2}(2x)\,dx =-\tfrac12\int_{u(0)}^{u(\frac{\pi}{4})}(1-u^{2})\,u^{2}\,du.

Passo 4

Os limites são
u(0)=\cos(0)=1, \quad u\bigl(\tfrac{\pi}{4}\bigr)=\cos\bigl(\tfrac{\pi}{2}\bigr)=0.
Expandindo e integrando:
(1-u^{2})\,u^{2}=u^{2}-u^{4},
-\tfrac12\int_{1}^{0}(u^{2}-u^{4})\,du =-\tfrac12\Bigl[\tfrac{u^{3}}{3}-\tfrac{u^{5}}{5}\Bigr]_{1}^{0}.

Passo 5

Calculando o valor numérico:
F(u)=-\tfrac12\Bigl(\tfrac{u^{3}}{3}-\tfrac{u^{5}}{5}\Bigr) =-\tfrac{u^{3}}{6}+\tfrac{u^{5}}{10},
\int_{0}^{\frac{\pi}{4}}\sin^{3}(2x)\cos^{2}(2x)\,dx =F\bigl(0\bigr)-F(1) =\bigl(0\bigr)-\Bigl(-\tfrac{1}{6}+\tfrac{1}{10}\Bigr) =\tfrac{1}{6}-\tfrac{1}{10} =\tfrac{1}{15}.

Resposta

\displaystyle \frac{1}{15}

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