Questão 8 - Calcule a integral das seguintes funções vetoriais:
\int_{-\pi}^{\pi / 2}\left(\cos ^{2}(t), \operatorname{sen}^{2}(t), t\right) d t.

Passo 1

O enunciado pede para calcular a integral vetorial
\int_{-\pi}^{\pi/2}\bigl(\cos^2 t,\;\sin^2 t,\;t\bigr)\,dt.
Como a integral vetorial é linear, vamos usar
\int_{a}^{b}\bigl(f(t),g(t),h(t)\bigr)\,dt =\Bigl(\int_{a}^{b}f(t)\,dt,\;\int_{a}^{b}g(t)\,dt,\;\int_{a}^{b}h(t)\,dt\Bigr),
e resolver cada componente separadamente.

Passo 2

Para a primeira componente, note que
\cos^2 t = \frac{1+\cos(2t)}{2}.
Assim,
I_1 = \int_{-\pi}^{\pi/2}\cos^2 t\,dt = \int_{-\pi}^{\pi/2}\frac{1+\cos(2t)}{2}\,dt = \left[\frac{t}{2} + \frac{\sin(2t)}{4}\right]_{-\pi}^{\pi/2}.
Ao avaliar:
\left.\frac{t}{2} + \frac{\sin(2t)}{4}\right|_{t=\pi/2} = \frac{\pi}{4} + 0, \quad \left.\frac{t}{2} + \frac{\sin(2t)}{4}\right|_{t=-\pi} = -\frac{\pi}{2} + 0,
logo
I_1 = \frac{\pi}{4} - \Bigl(-\frac{\pi}{2}\Bigr) = \frac{3\pi}{4}.

Passo 3

Para a segunda componente, use
\sin^2 t = \frac{1-\cos(2t)}{2}.
Então
I_2 = \int_{-\pi}^{\pi/2}\sin^2 t\,dt = \int_{-\pi}^{\pi/2}\frac{1-\cos(2t)}{2}\,dt = \left[\frac{t}{2} - \frac{\sin(2t)}{4}\right]_{-\pi}^{\pi/2}.
Ao avaliar:
\left.\frac{t}{2} - \frac{\sin(2t)}{4}\right|_{t=\pi/2} = \frac{\pi}{4} - 0, \quad \left.\frac{t}{2} - \frac{\sin(2t)}{4}\right|_{t=-\pi} = -\frac{\pi}{2} - 0,
logo
I_2 = \frac{\pi}{4} - \Bigl(-\frac{\pi}{2}\Bigr) = \frac{3\pi}{4}.

Passo 4

Para a terceira componente basta integrar t:
I_3 = \int_{-\pi}^{\pi/2} t\,dt = \left[\frac{t^2}{2}\right]_{-\pi}^{\pi/2} = \frac{(\pi/2)^2}{2} - \frac{(-\pi)^2}{2} = \frac{\pi^2}{8} - \frac{\pi^2}{2} = -\frac{3\pi^2}{8}.

Resposta

\int_{-\pi}^{\pi/2}\bigl(\cos^2 t,\sin^2 t,t\bigr)\,dt = \Bigl(\tfrac{3\pi}{4},\;\tfrac{3\pi}{4},\;-\tfrac{3\pi^2}{8}\Bigr).

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