Calcule as seguintes integrais indicando o método utilizado
\int_{0}^{\frac{\pi}{2}} \sin ^{4} \theta \cos ^{2} \theta d \theta

Passo 1

O enunciado pede para calcular
I = \int_{0}^{\frac{\pi}{2}} \sin^{4}\theta \,\cos^{2}\theta \,d\theta
usando um método adequado. Aqui vamos aplicar a fórmula que relaciona integrais de potências de seno e cosseno com a função Beta.

Passo 2

Lembramos que, para m>0 e n>0, vale
\int_{0}^{\frac{\pi}{2}}\sin^{m-1}\theta\,\cos^{n-1}\theta\,d\theta = \tfrac12\,B\bigl(\tfrac{m}{2},\tfrac{n}{2}\bigr),
onde
B(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}
é a função Beta.

Passo 3

No nosso caso temos \sin^{4}\theta\cos^{2}\theta=\sin^{m-1}\theta\cos^{n-1}\theta com
m-1=4\quad\Rightarrow\quad m=5, \qquad n-1=2\quad\Rightarrow\quad n=3.
Logo
I=\int_{0}^{\frac{\pi}{2}}\sin^{4}\theta\,\cos^{2}\theta\,d\theta = \tfrac12\,B\bigl(\tfrac{5}{2},\tfrac{3}{2}\bigr).

Passo 4

Expressamos B\bigl(\tfrac{5}{2},\tfrac{3}{2}\bigr) via \Gamma:
B\bigl(\tfrac{5}{2},\tfrac{3}{2}\bigr) = \frac{\Gamma(\tfrac{5}{2})\,\Gamma(\tfrac{3}{2})}{\Gamma(4)}.

Passo 5

Usamos os valores
\Gamma\bigl(\tfrac{5}{2}\bigr)=\tfrac{3}{2}\cdot\tfrac{1}{2}\,\sqrt{\pi}=\frac{3\sqrt{\pi}}{4}, \qquad \Gamma\bigl(\tfrac{3}{2}\bigr)=\tfrac{1}{2}\,\sqrt{\pi}, \qquad \Gamma(4)=3!=6.
Então
B\bigl(\tfrac{5}{2},\tfrac{3}{2}\bigr) = \frac{\tfrac{3\sqrt{\pi}}{4}\,\cdot\tfrac{\sqrt{\pi}}{2}}{6} = \frac{3\pi}{8}\,\frac{1}{6} = \frac{3\pi}{48} = \frac{\pi}{16}.

Passo 6

Finalmente
I = \tfrac12\,B\bigl(\tfrac{5}{2},\tfrac{3}{2}\bigr) = \tfrac12 \,\frac{\pi}{16} = \frac{\pi}{32}.

Resposta

\int_{0}^{\frac{\pi}{2}}\sin^{4}\theta\,\cos^{2}\theta\,d\theta = \frac{\pi}{32}.

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