Resolva as integrais:
\int_{0}^{\frac{\pi}{2}} \operatorname{sen}^{2}\left(\frac{x}{3}\right) d x

Passo 1

O enunciado pede para calcular a integral
\int_{0}^{\frac{\pi}{2}} \operatorname{sen}^{2}\bigl(\tfrac{x}{3}\bigr)\,dx.
O conteúdo principal é integração de funções trigonométricas usando identidades e mudança de variável. Vamos usar a substituição u=\tfrac{x}{3} e a identidade
\operatorname{sen}^{2}(u)=\frac{1-\cos(2u)}{2}.

Passo 2

Fazendo a substituição u=\tfrac{x}{3}, temos x=3u e dx=3\,du. Quando x vai de 0 a \tfrac{\pi}{2}, u vai de
u\bigl|_{x=0}=0 \quad\text{até}\quad u\bigl|_{x=\frac{\pi}{2}}=\frac{\pi}{6}.é
Logo,
\int_{0}^{\frac{\pi}{2}} \operatorname{sen}^{2}\bigl(\tfrac{x}{3}\bigr)\,dx =3\int_{0}^{\frac{\pi}{6}} \operatorname{sen}^{2}(u)\,du.

Passo 3

Aplicamos a identidade \operatorname{sen}^{2}(u)=\tfrac{1-\cos(2u)}{2}:
3\int_{0}^{\frac{\pi}{6}} \operatorname{sen}^{2}(u)\,du =3\int_{0}^{\frac{\pi}{6}} \frac{1-\cos(2u)}{2}\,du =\frac{3}{2}\int_{0}^{\frac{\pi}{6}}\bigl(1-\cos(2u)\bigr)\,du.

Passo 4

Integramos termo a termo:
\int\bigl(1-\cos(2u)\bigr)\,du =\int 1\,du-\int\cos(2u)\,du =u-\tfrac{1}{2}\operatorname{sen}(2u).
Portanto,
\frac{3}{2}\biggl[u-\tfrac{1}{2}\operatorname{sen}(2u)\biggr]_{0}^{\frac{\pi}{6}} =\frac{3}{2}\Bigl(\bigl(\tfrac{\pi}{6}-\tfrac{1}{2}\operatorname{sen}\tfrac{\pi}{3}\bigr)-0\Bigr).

Passo 5

Sabendo que \operatorname{sen}\tfrac{\pi}{3}=\tfrac{\sqrt{3}}{2}, obtemos
\frac{3}{2}\Bigl(\frac{\pi}{6}-\frac{1}{2}\cdot\frac{\sqrt{3}}{2}\Bigr) =\frac{3}{2}\Bigl(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\Bigr) =\frac{3\pi}{12}-\frac{3\sqrt{3}}{8} =\frac{\pi}{4}-\frac{3\sqrt{3}}{8}.

Resposta

\int_{0}^{\frac{\pi}{2}} \operatorname{sen}^{2}\bigl(\tfrac{x}{3}\bigr)\,dx =\frac{\pi}{4}-\frac{3\sqrt{3}}{8}.

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