Calcule a integral.
\int_{0}^{\pi} \operatorname{sen}^{2} t \cos ^{4} t d t

Passo 1

O enunciado pede para calcular
\int_{0}^{\pi}\operatorname{sen}^{2}t\;\cos^{4}t\;dt.
O conteúdo principal aqui é usar simetria da função no intervalo [0,\pi] e a fórmula de Beta/Gamma para integrais do tipo
\int_{0}^{\frac{\pi}{2}}\operatorname{sen}^{p}t\;\cos^{q}t\;dt.

Passo 2

Note que f(t)=\operatorname{sen}^{2}t\cos^{4}t satisfaz f(\pi-t)=\operatorname{sen}^{2}(\pi-t)\,\cos^{4}(\pi-t)=\operatorname{sen}^{2}t\,(-\cos t)^{4}=\operatorname{sen}^{2}t\cos^{4}t, ou seja, é par em torno de \tfrac{\pi}{2}. Logo
\int_{0}^{\pi}\operatorname{sen}^{2}t\cos^{4}t\;dt =2\int_{0}^{\frac{\pi}{2}}\operatorname{sen}^{2}t\cos^{4}t\;dt.

Passo 3

Usamos a fórmula de Beta: para p>-1 e q>-1,
\int_{0}^{\frac{\pi}{2}}\operatorname{sen}^{p}t\;\cos^{q}t\;dt =\tfrac12\,B\bigl(\tfrac{p+1}2,\tfrac{q+1}2\bigr)
com
B(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}.
Aqui p=2 e q=4, portanto
\int_{0}^{\frac{\pi}{2}}\operatorname{sen}^{2}t\cos^{4}t\;dt =\tfrac12\,B\bigl(\tfrac{3}{2},\tfrac{5}{2}\bigr).

Passo 4

Calculemos
\Gamma\bigl(\tfrac32\bigr)=\frac{\sqrt\pi}{2},\quad \Gamma\bigl(\tfrac52\bigr)=\frac{3\sqrt\pi}{4},\quad \Gamma\bigl(\tfrac32+\tfrac52\bigr)=\Gamma(4)=6.
Logo
B\bigl(\tfrac32,\tfrac52\bigr) =\frac{\frac{\sqrt\pi}{2}\,\frac{3\sqrt\pi}{4}}{6} =\frac{3\pi/8}{6} =\frac{3\pi}{48} =\frac{\pi}{16}.
Assim
\int_{0}^{\frac{\pi}{2}}\operatorname{sen}^{2}t\cos^{4}t\;dt =\tfrac12\cdot\frac{\pi}{16} =\frac{\pi}{32}.

Passo 5

Voltando à integral original,
\int_{0}^{\pi}\operatorname{sen}^{2}t\cos^{4}t\;dt =2\cdot\frac{\pi}{32} =\frac{\pi}{16}.

Resposta

\displaystyle \frac{\pi}{16}

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