1. Calcule as seguintes integrais:
    a) \int \cos ^{5} x \operatorname{sen} x d x.
    b) \int_{0}^{\pi / 6} \cos ^{2} x d x.
    c) \int \operatorname{sen}^{2} x \cos ^{3} x d x.
    d) \int \operatorname{sen}^{2} 2 \theta \cos 2 \theta d \theta.
    е) \int_{0}^{\pi / 3} \cos ^{3} x d x
    g) \int \operatorname{sen}^{2} x \cos 2 x d x.

Passo 1

O enunciado pede o cálculo de diversas integrais de funções trigonométricas com potências de \operatorname{sen} x, \cos x e argumentos múltiplos. Para resolver usaremos principalmente:
  • Substituições u quando aparecer \operatorname{sen} x ou \cos x junto da sua derivada.
  • Identidades de redução de potência, por exemplo
    \cos^2x=\tfrac{1+\cos2x}{2} e \operatorname{sen}^2x=\tfrac{1-\cos2x}{2}.
  • Avaliação de limites para integrais definidas usando as antiderivadas obtidas.

Passo 2

Calcule \displaystyle \int \cos^5 x\,\operatorname{sen} x\,dx.
Note que \frac{d}{dx}(\cos x)=-\operatorname{sen} x. Faça u=\cos x, du=-\operatorname{sen} x\,dx. Então
\int \cos^5 x\,\operatorname{sen} x\,dx = -\int u^5\,du = -\frac{u^6}{6}+C = -\frac{\cos^6 x}{6}+C.

Passo 3

Calcule \displaystyle \int_{0}^{\pi/6}\cos^2 x\,dx.
Use \cos^2x=\tfrac{1+\cos2x}{2}. Assim
\int_{0}^{\pi/6}\cos^2 x\,dx = \frac12\int_{0}^{\pi/6}(1+\cos2x)\,dx = \frac12\Bigl[x + \tfrac{\sin2x}{2}\Bigr]_{0}^{\pi/6} = \frac{\pi}{12} + \frac{\sin(\pi/3)}{4} = \frac{\pi}{12} + \frac{\sqrt3}{8}.

Passo 4

Calcule \displaystyle \int \operatorname{sen}^2 x\,\cos^3 x\,dx.
Escreva \cos^3x=\cos x\,(1-\operatorname{sen}^2x) e faça u=\operatorname{sen} x, du=\cos x\,dx. Logo
\int \operatorname{sen}^2 x\,\cos^3 x\,dx = \int u^2(1-u^2)\,du = \int(u^2 - u^4)\,du = \frac{u^3}{3}-\frac{u^5}{5}+C = \frac{\operatorname{sen}^3 x}{3}-\frac{\operatorname{sen}^5 x}{5}+C.

Passo 5

Calcule \displaystyle \int \operatorname{sen}^2(2\theta)\,\cos2\theta\,d\theta.
Faça u=\sen2\theta, du=2\cos2\theta\,d\theta. Então
\int \operatorname{sen}^2(2\theta)\cos2\theta\,d\theta = \frac12\int u^2\,du = \frac{u^3}{6}+C = \frac{\operatorname{sen}^3(2\theta)}{6}+C.

Passo 6

Calcule \displaystyle \int_{0}^{\pi/3}\cos^3 x\,dx.
Use \cos^3x=\cos x\,(1-\operatorname{sen}^2x) com u=\operatorname{sen} x, du=\cos x\,dx. A antiderivada é
\sin x - \frac{\sin^3 x}{3}
e avaliando de 0 a \pi/3 resulta em
\Bigl(\tfrac{\sqrt3}{2} - \tfrac{(\sqrt3/2)^3}{3}\Bigr) = \frac{\sqrt3}{2}-\frac{\sqrt3}{8} = \frac{3\sqrt3}{8}.

Passo 7

Calcule \displaystyle \int \operatorname{sen}^2 x\,\cos2x\,dx.
Use \operatorname{sen}^2x=\tfrac{1-\cos2x}{2}, então
\operatorname{sen}^2x\cos2x = \frac12(\cos2x - \cos^2 2x), \quad \cos^2 2x=\frac{1+\cos4x}{2}.
Logo
\int \operatorname{sen}^2 x\cos2x\,dx = \frac12\int\cos2x\,dx - \frac12\int\cos^2 2x\,dx = \frac{\sin2x}{4} - \frac14\Bigl(x + \frac{\sin4x}{4}\Bigr) + C = \frac{\sin2x}{4}-\frac{x}{4}-\frac{\sin4x}{16}+C.

Resposta

a) \displaystyle -\frac{\cos^6 x}{6}+C.
b) \displaystyle \frac{\pi}{12}+\frac{\sqrt3}{8}.
c) \displaystyle \frac{\operatorname{sen}^3 x}{3}-\frac{\operatorname{sen}^5 x}{5}+C.
d) \displaystyle \frac{\operatorname{sen}^3(2\theta)}{6}+C.
e) \displaystyle \frac{3\sqrt3}{8}.
g) \displaystyle \frac{\sin2x}{4}-\frac{x}{4}-\frac{\sin4x}{16}+C.

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