Utilize a mudança de variável u=\operatorname{tg}\left(\frac{x}{2}\right) para calcular as integrais seguintes:
\int_{0}^{\frac{\pi}{2}} \frac{1}{2+\operatorname{sen} x} d x.

Passo 1

O enunciado pede calcular
\int_{0}^{\frac{\pi}{2}}\frac{1}{2+\operatorname{sen} x}\,dx
usando a mudança de variável de Weierstrass u=\tan\!\bigl(\tfrac{x}{2}\bigr). Para isso, vamos precisar das fórmulas \operatorname{sen} x=\tfrac{2u}{1+u^2}, dx=\tfrac{2\,du}{1+u^2}
e dos novos limites correspondentes a x=0 e x=\tfrac{\pi}{2}.

Passo 2

Aplicando u=\tan\!\bigl(\tfrac{x}{2}\bigr) temos
\operatorname{sen} x = \frac{2u}{1+u^2}, \quad dx = \frac{2\,du}{1+u^2}.
Quando x=0 vem u=\tan0=0, e quando x=\tfrac{\pi}{2} vem u=\tan\!\bigl(\tfrac{\pi}{4}\bigr)=1.

Passo 3

Substituindo na integral original:
2+\operatorname{sen} x =2+\frac{2u}{1+u^2} =\frac{2(1+u^2)+2u}{1+u^2} =\frac{2(u^2+u+1)}{1+u^2}.
Logo
\frac{1}{2+\operatorname{sen} x} =\frac{1+u^2}{2(u^2+u+1)},
e
\frac{1}{2+\operatorname{sen} x}\,dx =\frac{1+u^2}{2(u^2+u+1)}\;\frac{2\,du}{1+u^2} =\frac{du}{u^2+u+1}.
Portanto a integral vira
\int_{0}^{\frac{\pi}{2}}\frac{1}{2+\operatorname{sen} x}\,dx =\int_{0}^{1}\frac{du}{u^2+u+1}.

Passo 4

Para integrar \displaystyle\int\frac{du}{u^2+u+1}, complete o quadrado no denominador:
u^2+u+1 =\Bigl(u+\tfrac12\Bigr)^2+\tfrac34.
Então
\int\frac{du}{u^2+u+1} =\int\frac{du}{\bigl(u+\tfrac12\bigr)^2+(\tfrac{\sqrt{3}}{2})^2} =\frac{2}{\sqrt{3}} \arctan\!\Bigl(\frac{u+\tfrac12}{\tfrac{\sqrt3}{2}}\Bigr) +C.

Passo 5

Avalia dos limites 0 a 1:
I =\frac{2}{\sqrt3}\Bigl[ \arctan\!\bigl(\tfrac{1+\tfrac12}{\tfrac{\sqrt3}{2}}\bigr) -\arctan\!\bigl(\tfrac{0+\tfrac12}{\tfrac{\sqrt3}{2}}\bigr) \Bigr] =\frac{2}{\sqrt3}\Bigl[\arctan(\sqrt3)-\arctan\bigl(\tfrac1{\sqrt3}\bigr)\Bigr].
Como \arctan(\sqrt3)=\tfrac{\pi}{3} e \arctan(\tfrac1{\sqrt3})=\tfrac{\pi}{6}, obtemos
I =\frac{2}{\sqrt3}\Bigl(\tfrac{\pi}{3}-\tfrac{\pi}{6}\Bigr) =\frac{2}{\sqrt3}\;\frac{\pi}{6} =\frac{\pi}{3\sqrt3}.

Resposta

\int_{0}^{\frac{\pi}{2}}\frac{1}{2+\operatorname{sen} x}\,dx =\frac{\pi}{3\sqrt3}.

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