( \mathbf{3 , 0} ptos) Calcule as integrais usando funções Gamma e Beta
\int_{0}^{\frac{\pi}{2}} \sin ^{4} \theta \cos ^{4} \theta d \theta

Passo 1

O enunciado pede para calcular a integral
I = \int_{0}^{\frac{\pi}{2}} \sin^{4}\theta\,\cos^{4}\theta\,d\theta
usando as funções Beta e Gamma. Para isso, vamos usar a fórmula padrão
\int_{0}^{\frac{\pi}{2}}\sin^{m-1}\theta\cos^{n-1}\theta\,d\theta = \tfrac12\,B\bigl(\tfrac m2,\tfrac n2\bigr)
e a relação
B(a,b)=\frac{\Gamma(a)\,\Gamma(b)}{\Gamma(a+b)}.

Passo 2

Na nossa integral temos \sin^{4}\theta=\sin^{(5-1)}\theta e \cos^{4}\theta=\cos^{(5-1)}\theta, ou seja, m-1=4 e n-1=4, logo
m=5,\quad n=5
e portanto
I = \tfrac12\,B\bigl(\tfrac{5}{2},\tfrac{5}{2}\bigr).

Passo 3

Usando a definição de Beta em termos de Gamma,
B\bigl(\tfrac{5}{2},\tfrac{5}{2}\bigr) = \frac{\Gamma\bigl(\tfrac{5}{2}\bigr)\,\Gamma\bigl(\tfrac{5}{2}\bigr)}{\Gamma(5)}.
Sabemos que \Gamma(5)=4!=24 e que
\Gamma\bigl(\tfrac{5}{2}\bigr) =\frac{3}{2}\cdot\frac{1}{2}\,\Gamma\bigl(\tfrac{1}{2}\bigr) =\frac{3}{4}\,\sqrt{\pi}.
Logo
\Gamma\bigl(\tfrac{5}{2}\bigr)^{2} =\bigl(\tfrac{3}{4}\sqrt{\pi}\bigr)^{2} =\frac{9\pi}{16}.

Passo 4

Substituindo em B e depois em I, temos
B\bigl(\tfrac{5}{2},\tfrac{5}{2}\bigr) = \frac{\tfrac{9\pi}{16}}{24} = \frac{9\pi}{384} = \frac{3\pi}{128},
logo
I = \tfrac12 \cdot \frac{3\pi}{128} = \frac{3\pi}{256}.

Resposta

\int_{0}^{\frac{\pi}{2}} \sin^{4}\theta\,\cos^{4}\theta\,d\theta = \frac{3\pi}{256}.

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