Calcule as integrais nos
\int_{0}^{\pi / 3}[(\operatorname{sec} t \operatorname{tg} t) \mathbf{i}+(\operatorname{tg} t) \mathbf{j}+(2 \operatorname{sen} t \cos t) \mathbf{k}]

Passo 1

O enunciado pede para calcular a integral vetorial
\int_{0}^{\pi/3}\bigl[(\operatorname{sec} t\tan t)\,\mathbf{i}+(\tan t)\,\mathbf{j}+(2\sin t\cos t)\,\mathbf{k}\bigr]\,dt.
O conteúdo aqui é integração de funções trigonométricas elementares, usando derivadas conhecidas \frac{d}{dt}\operatorname{sec} t=\operatorname{sec} t\tan t e identidades 2\sin t\cos t=\sin2t.

Passo 2

Componente \mathbf{i}:
\int_{0}^{\pi/3}\operatorname{sec} t\tan t\,dt
Como \frac{d}{dt}\operatorname{sec} t=\operatorname{sec} t\tan t, temos
\int\operatorname{sec} t\tan t\,dt=\operatorname{sec} t,
logo
\int_{0}^{\pi/3}\operatorname{sec} t\tan t\,dt =\bigl[\operatorname{sec} t\bigr]_{0}^{\pi/3} =\operatorname{sec}\bigl(\tfrac{\pi}{3}\bigr)-\operatorname{sec}(0) =2-1=1.

Passo 3

Componente \mathbf{j}:
\int_{0}^{\pi/3}\tan t\,dt.
Usamos \int\tan t\,dt=-\ln\bigl|\cos t\bigr|, então
\int_{0}^{\pi/3}\tan t\,dt =-\bigl[\ln\lvert\cos t\rvert\bigr]_{0}^{\pi/3} =-\ln\bigl(\tfrac12\bigr)+\ln(1) =\ln2.

Passo 4

Componente \mathbf{k}:
\int_{0}^{\pi/3}2\sin t\cos t\,dt.
Como 2\sin t\cos t=\sin2t,
\int\sin2t\,dt=-\tfrac12\cos2t,
então
\int_{0}^{\pi/3}2\sin t\cos t\,dt =\left[-\tfrac12\cos2t\right]_{0}^{\pi/3} =\Bigl(-\tfrac12\cos\tfrac{2\pi}{3}\Bigr)-\Bigl(-\tfrac12\cos0\Bigr) =\tfrac14+\tfrac12 =\tfrac{3}{4}.

Passo 5

Reunindo as três componentes, o resultado da integral vetorial é
1\,\mathbf{i}+\ln2\,\mathbf{j}+\tfrac{3}{4}\,\mathbf{k}.

Resposta

\mathbf{i}+\ln2\,\mathbf{j}+\tfrac{3}{4}\,\mathbf{k}.

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