Calcule as integrais trigonométricas indefinidas mostradas abaixo.
\int \operatorname{sen}^{3}(1-2 \theta) \cos ^{3}(1-2 \theta) d \theta

Passo 1

O enunciado pede para calcular
\int \operatorname{sen}^{3}(1-2\theta)\,\cos^{3}(1-2\theta)\,d\theta.
O conteúdo envolve integração de potências ímpares de funções trigonométricas. Vamos usar mudança de variável para simplificar o argumento e depois aplicar identidades de ângulo duplo para reduzir a potência.

Passo 2

Fazemos a substituição
\varphi = 1 - 2\theta,
logo
d\varphi = -2\,d\theta \quad\Longrightarrow\quad d\theta = -\tfrac12\,d\varphi.
A integral passa a ser
\int \operatorname{sen}^{3}(1-2\theta)\cos^{3}(1-2\theta)\,d\theta = -\tfrac12 \int \operatorname{sen}^{3}\varphi\,\cos^{3}\varphi\,d\varphi.

Passo 3

Usamos a identidade de ângulo duplo para combinar potências:
\operatorname{sen}^{3}\varphi\,\cos^{3}\varphi = (\operatorname{sen}\varphi\,\cos\varphi)\,(\operatorname{sen}^{2}\varphi\,\cos^{2}\varphi) = \bigl(\tfrac12\operatorname{sen}(2\varphi)\bigr)\,\bigl(\operatorname{sen}^{2}\varphi\,\cos^{2}\varphi\bigr).
E como
\operatorname{sen}^{2}\varphi\,\cos^{2}\varphi = \bigl(\tfrac12\operatorname{sen}(2\varphi)\bigr)^{2} = \tfrac14\,\operatorname{sen}^{2}(2\varphi),
segue que
\operatorname{sen}^{3}\varphi\,\cos^{3}\varphi = \tfrac12\cdot\tfrac14\,\operatorname{sen}(2\varphi)\,\operatorname{sen}^{2}(2\varphi) = \tfrac18\,\operatorname{sen}^{3}(2\varphi).
Portanto a integral fica
-\tfrac12\int \operatorname{sen}^{3}\varphi\,\cos^{3}\varphi\,d\varphi = -\tfrac12\cdot\tfrac18\int \operatorname{sen}^{3}(2\varphi)\,d\varphi = -\tfrac1{16}\int \operatorname{sen}^{3}(2\varphi)\,d\varphi.

Passo 4

Para integrar \int \operatorname{sen}^{3}(2\varphi)\,d\varphi, usamos u=2\varphi, du=2\,d\varphi ou d\varphi=\tfrac12du:
\int \operatorname{sen}^{3}(2\varphi)\,d\varphi = \tfrac12\int \operatorname{sen}^{3}u\,du.
E sabemos que
\int \operatorname{sen}^{3}u\,du = \int \operatorname{sen} u\,(1-\cos^{2}u)\,du = -\cos u + \tfrac13\cos^{3}u + C.
Logo
\int \operatorname{sen}^{3}(2\varphi)\,d\varphi = \tfrac12\Bigl[-\cos u + \tfrac13\cos^{3}u\Bigr] + C = -\tfrac12\cos(2\varphi) + \tfrac16\cos^{3}(2\varphi) + C.

Passo 5

Voltando à constante -\tfrac1{16}, temos
-\tfrac1{16}\int \operatorname{sen}^{3}(2\varphi)\,d\varphi = -\tfrac1{16}\Bigl[-\tfrac12\cos(2\varphi) + \tfrac16\cos^{3}(2\varphi)\Bigr] + C = \tfrac1{32}\cos(2\varphi) - \tfrac1{96}\cos^{3}(2\varphi) + C.

Passo 6

Substituímos \varphi=1-2\theta, de modo que 2\varphi=2-4\theta. Assim a primitiva original é
\boxed{ \int \operatorname{sen}^{3}(1-2\theta)\cos^{3}(1-2\theta)\,d\theta = \tfrac1{32}\cos\bigl(2-4\theta\bigr)\;-\;\tfrac1{96}\cos^{3}\bigl(2-4\theta\bigr) + C. }

Resposta

\int \operatorname{sen}^{3}(1-2\theta)\cos^{3}(1-2\theta)\,d\theta = \tfrac1{32}\cos\bigl(2-4\theta\bigr)\;-\;\tfrac1{96}\cos^{3}\bigl(2-4\theta\bigr) + C.

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