Calcule as integrais nos
\int_{-\pi / 4}^{\pi / 4}\left[(\operatorname{sen} t) \mathbf{i}+(1+\cos t) \mathbf{j}+\left(\operatorname{sec} ^{2} t\right) \mathbf{k}\right] d t

Passo 1

O enunciado pede a integral vetorial de \mathbf{F}(t)=\bigl(\sin t\bigr)\mathbf{i}+\bigl(1+\cos t\bigr)\mathbf{j}+\bigl(\operatorname{sec}^2 t\bigr)\mathbf{k} no intervalo \bigl[-\tfrac{\pi}{4},\tfrac{\pi}{4}\bigr]. Vamos calcular componente a componente.

Passo 2

Para a componente em \mathbf{i}:
\int_{-\pi/4}^{\pi/4}\sin t\,dt = \bigl[-\cos t\bigr]_{-\pi/4}^{\pi/4} = -\cos\bigl(\tfrac{\pi}{4}\bigr)+\cos\bigl(-\tfrac{\pi}{4}\bigr) = -\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2} = 0.

Passo 3

Para a componente em \mathbf{j}:
\int_{-\pi/4}^{\pi/4}(1+\cos t)\,dt = \bigl[t+\sin t\bigr]_{-\pi/4}^{\pi/4} = \Bigl(\tfrac{\pi}{4}+\tfrac{\sqrt2}{2}\Bigr) - \Bigl(-\tfrac{\pi}{4}-\tfrac{\sqrt2}{2}\Bigr) = \tfrac{\pi}{2} + \sqrt2.

Passo 4

Para a componente em \mathbf{k}:
\int_{-\pi/4}^{\pi/4}\operatorname{sec}^2 t\,dt = \bigl[\tan t\bigr]_{-\pi/4}^{\pi/4} = \tan\bigl(\tfrac{\pi}{4}\bigr)-\tan\bigl(-\tfrac{\pi}{4}\bigr) = 1 - (-1) = 2.

Passo 5

Reunindo as componentes, a integral vetorial é
0\,\mathbf{i} + \Bigl(\tfrac{\pi}{2} + \sqrt2\Bigr)\mathbf{j} + 2\,\mathbf{k}.

Resposta

0\,\mathbf{i} + \bigl(\tfrac{\pi}{2} + \sqrt2\bigr)\mathbf{j} + 2\,\mathbf{k}

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