Mostre que:
\int_{0}^{\frac{\pi}{3}} e^{x} \operatorname{sen}(x) d x=\frac{1}{2}+\frac{1}{4}(\sqrt{3}-1) e^{\frac{\pi}{3}}

Passo 1

O enunciado pede para calcular a integral definida
\int_{0}^{\frac{\pi}{3}} e^{x}\operatorname{sen}(x)\,dx
usando técnicas de integração por partes. Vamos primeiro encontrar a primitiva geral de
\displaystyle I(x)=\int e^{x}\operatorname{sen}(x)\,dx.

Passo 2

Defina
I(x)=\int e^{x}\operatorname{sen}(x)\,dx.
Aplique integração por partes escolhendo
\displaystyle u=e^{x}, \,dv=\operatorname{sen}(x)\,dx. Então
\,du=e^{x}\,dx e \,v=-\cos(x), logo
I(x) =\;u\,v-\int v\,du =-e^{x}\cos(x)+\int e^{x}\cos(x)\,dx.
Chame
\displaystyle J(x)=\int e^{x}\cos(x)\,dx e aplique partes em J(x) com
\displaystyle u=e^{x}, \,dv=\cos(x)\,dx. Assim
\,du=e^{x}\,dx, \,v=\operatorname{sen}(x) e
J(x) =e^{x}\operatorname{sen}(x)-\int e^{x}\operatorname{sen}(x)\,dx =e^{x}\operatorname{sen}(x)-I(x).
Substituindo J(x) em I(x) obtém-se
I(x)=-e^{x}\cos(x)+J(x) =-e^{x}\cos(x)+e^{x}\operatorname{sen}(x)-I(x),
ou seja
2\,I(x)=e^{x}\bigl(\operatorname{sen}(x)-\cos(x)\bigr) \quad\Longrightarrow\quad I(x)=\frac{e^{x}}{2}\bigl(\operatorname{sen}(x)-\cos(x)\bigr).

Passo 3

Agora avalie a primitiva nos limites de 0 a \pi/3:
\int_{0}^{\frac{\pi}{3}}e^{x}\operatorname{sen}(x)\,dx =\left.\frac{e^{x}}{2}\bigl(\operatorname{sen}(x)-\cos(x)\bigr)\right|_{0}^{\frac{\pi}{3}}.
Para x=\tfrac{\pi}{3}:
\operatorname{sen}(\tfrac{\pi}{3})=\tfrac{\sqrt{3}}{2}, \cos(\tfrac{\pi}{3})=\tfrac{1}{2}, então
\frac{e^{\frac{\pi}{3}}}{2}\Bigl(\tfrac{\sqrt{3}}{2}-\tfrac{1}{2}\Bigr) =\frac{e^{\frac{\pi}{3}}}{2}\cdot\frac{\sqrt{3}-1}{2} =\frac{\sqrt{3}-1}{4}\,e^{\frac{\pi}{3}}.
Para x=0:
\operatorname{sen}(0)=0, \cos(0)=1, então
\frac{e^{0}}{2}\bigl(0-1\bigr) =-\frac{1}{2}.
Subtraindo, vem
\int_{0}^{\frac{\pi}{3}}e^{x}\operatorname{sen}(x)\,dx =\frac{\sqrt{3}-1}{4}\,e^{\frac{\pi}{3}} \;-\;\bigl(-\tfrac12\bigr) =\frac12+\frac{\sqrt{3}-1}{4}\,e^{\frac{\pi}{3}}.

Resposta

\int_{0}^{\frac{\pi}{3}} e^{x}\operatorname{sen}(x)\,dx =\frac12+\frac{1}{4}\bigl(\sqrt{3}-1\bigr)e^{\frac{\pi}{3}}.

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