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.
e) \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 para calcular várias integrais envolvendo potências de seno e cosseno, algumas definidas e outras indefinidas. O conteúdo principal aqui são duas técnicas:
  1. substituição simples (por exemplo u=\cos x ou u=\sin x),
  2. uso de identidades de redução de potência \cos^2x=(1+\cos2x)/2, \sin^2x=(1-\cos2x)/2.

Passo 2

Calcular \displaystyle \int \cos^5x\,\operatorname{sen} x\,dx.
Escolhemos u=\cos x, daí du=-\operatorname{sen} x\,dx e
\int \cos^5x\,\operatorname{sen} x\,dx =\int u^5(-du) =-\int u^5\,du =-\frac{u^6}{6}+C =-\frac{\cos^6x}{6}+C.

Passo 3

Calcular \displaystyle \int_{0}^{\pi/6}\cos^2x\,dx.
Usamos \cos^2x=(1+\cos2x)/2, então
\int_{0}^{\pi/6}\cos^2x\,dx =\frac12\int_{0}^{\pi/6}\bigl(1+\cos2x\bigr)dx =\frac12\Bigl[x+\frac{\sen2x}{2}\Bigr]_{0}^{\pi/6} =\frac12\Bigl(\frac{\pi}{6}+\frac{\operatorname{sen}(\pi/3)}{2}\Bigr) =\frac{\pi}{12}+\frac{\sqrt3}{8}.

Passo 4

Calcular \displaystyle \int \operatorname{sen}^2x\,\cos^3x\,dx.
Escrevemos \operatorname{sen}^2x=\bigl(\operatorname{sen} x\bigr)^2 e fazemos u=\operatorname{sen} x, du=\cos x\,dx. Como \cos^3x=\cos^2x\cdot\cos x=(1-\operatorname{sen}^2x)\cos x,
\int \operatorname{sen}^2x\,\cos^3x\,dx =\int u^2(1-u^2)\,du =\int\bigl(u^2-u^4\bigr)du =\frac{u^3}{3}-\frac{u^5}{5}+C =\frac{\operatorname{sen}^3x}{3}-\frac{\operatorname{sen}^5x}{5}+C.

Passo 5

Calcular \displaystyle \int \operatorname{sen}^2(2\theta)\,\cos2\theta\,d\theta.
Fazemos u=\sen2\theta, então du=2\cos2\theta\,d\theta ou \cos2\theta\,d\theta=du/2.
\int \operatorname{sen}^2(2\theta)\cos2\theta\,d\theta =\int u^2\frac{du}{2} =\frac{1}{2}\int u^2\,du =\frac{u^3}{6}+C =\frac{\operatorname{sen}^3(2\theta)}{6}+C.

Passo 6

Calcular \displaystyle \int_{0}^{\pi/3}\cos^3x\,dx.
Escrevemos \cos^3x=\cos x\,(1-\operatorname{sen}^2x) e tomamos u=\operatorname{sen} x, du=\cos x\,dx.
\int\cos^3x\,dx =\int(1-u^2)\,du =u-\frac{u^3}{3}+C =\operatorname{sen} x-\frac{\operatorname{sen}^3x}{3}+C.
No intervalo:
\int_{0}^{\pi/3}\cos^3x\,dx =\Bigl[\operatorname{sen} x-\frac{\operatorname{sen}^3x}{3}\Bigr]_{0}^{\pi/3} =\Bigl(\frac{\sqrt3}{2}-\frac{(\sqrt3/2)^3}{3}\Bigr)-0 =\frac{\sqrt3}{2}-\frac{3\sqrt3/8}{3} =\frac{3\sqrt3}{8}.

Passo 7

Calcular \displaystyle \int \operatorname{sen}^2x\,\cos2x\,dx.
Usamos \operatorname{sen}^2x=(1-\cos2x)/2, então
\operatorname{sen}^2x\,\cos2x =\frac{1}{2}(1-\cos2x)\cos2x =\frac12\cos2x-\frac12\cos^2 2x =\frac12\cos2x-\frac14-\frac14\cos4x.
Integrando termo a termo,
\int\operatorname{sen}^2x\,\cos2x\,dx =\int\Bigl(-\tfrac14+\tfrac12\cos2x-\tfrac14\cos4x\Bigr)dx =-\frac{x}{4}+\frac{\sen2x}{4}-\frac{\sen4x}{16}+C.

Resposta

a) \displaystyle -\frac{\cos^6x}{6}+C
b) \displaystyle \frac{\pi}{12}+\frac{\sqrt3}{8}
c) \displaystyle \frac{\operatorname{sen}^3x}{3}-\frac{\operatorname{sen}^5x}{5}+C
d) \displaystyle \frac{\operatorname{sen}^3(2\theta)}{6}+C
e) \displaystyle \frac{3\sqrt3}{8}
g) \displaystyle -\frac{x}{4}+\frac{\sen2x}{4}-\frac{\sen4x}{16}+C

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