Resolva as integrais definidas e indefinidas abaixo:
\int_{-\pi / 3}^{\pi / 4}(\cos (x)+2 \operatorname{sen}(x)) d x

Passo 1

O enunciado pede para calcular a integral definida
\int_{-\pi/3}^{\pi/4}\bigl(\cos(x)+2\operatorname{sen}(x)\bigr)\,dx.
O conteúdo aqui é “integrais definidas e indefinidas”, ou seja, vamos primeiro achar a antiderivada da função e depois aplicar o Teorema Fundamental do Cálculo para avaliar nos limites.

Passo 2

Vamos obter a integral indefinida
\int\bigl(\cos(x)+2\operatorname{sen}(x)\bigr)\,dx.
Sabemos que
\int\cos(x)\,dx=\operatorname{sen}(x), \quad \int\operatorname{sen}(x)\,dx=-\cos(x).
Logo,
\int2\operatorname{sen}(x)\,dx =2\int\operatorname{sen}(x)\,dx =-2\cos(x).
Portanto a antiderivada é
F(x)=\operatorname{sen}(x)-2\cos(x)+C.

Passo 3

Agora aplicamos o Teorema Fundamental do Cálculo:
\int_{-\pi/3}^{\pi/4}\bigl(\cos(x)+2\operatorname{sen}(x)\bigr)\,dx =F\bigl(\tfrac{\pi}{4}\bigr)-F\bigl(-\tfrac{\pi}{3}\bigr).

Passo 4

Calculamos cada termo:
Para x=\tfrac{\pi}{4},
\operatorname{sen}\!\bigl(\tfrac{\pi}{4}\bigr)=\tfrac{\sqrt{2}}{2}, \quad \cos\!\bigl(\tfrac{\pi}{4}\bigr)=\tfrac{\sqrt{2}}{2},
então
F\bigl(\tfrac{\pi}{4}\bigr) =\tfrac{\sqrt{2}}{2}-2\cdot\tfrac{\sqrt{2}}{2} =-\tfrac{\sqrt{2}}{2}.
Para x=-\tfrac{\pi}{3},
\operatorname{sen}\!\bigl(-\tfrac{\pi}{3}\bigr)=-\tfrac{\sqrt{3}}{2}, \quad \cos\!\bigl(-\tfrac{\pi}{3}\bigr)=\tfrac{1}{2},
então
F\bigl(-\tfrac{\pi}{3}\bigr) =-\tfrac{\sqrt{3}}{2}-2\cdot\tfrac{1}{2} =-\tfrac{\sqrt{3}}{2}-1.

Passo 5

Subtraindo, obtemos
F\bigl(\tfrac{\pi}{4}\bigr)-F\bigl(-\tfrac{\pi}{3}\bigr) =-\tfrac{\sqrt{2}}{2} -\bigl(-\tfrac{\sqrt{3}}{2}-1\bigr) =1+\frac{\sqrt{3}-\sqrt{2}}{2}.

Resposta

1+\dfrac{\sqrt{3}-\sqrt{2}}{2}

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