Calcular as seguintes integrais:
\int_{0}^{\pi} \int_{-1}^{1}|y-\cos (x)| d y d x.

Passo 1

O enunciado pede para calcular a integral dupla
\int_{0}^{\pi}\int_{-1}^{1}\bigl|y-\cos(x)\bigr|\,dy\,dx.
Este problema envolve valor absoluto e integração por partes em y, dependendo do ponto onde o termo dentro do módulo muda de sinal, isto é, y=\cos(x).

Passo 2

Para cada valor fixo de x\in[0,\pi], fazemos
\int_{-1}^{1}\bigl|y-\cos(x)\bigr|\,dy =\int_{-1}^{\cos(x)}\bigl(\cos(x)-y\bigr)\,dy +\int_{\cos(x)}^{1}\bigl(y-\cos(x)\bigr)\,dy,
pois para y<\cos(x) temos y-\cos(x)<0 e para y>\cos(x) temos y-\cos(x)>0.

Passo 3

Calculamos as duas integrais em y:
  1. \displaystyle \int_{-1}^{\cos(x)}\bigl(\cos(x)-y\bigr)\,dy
=\Bigl[\cos(x)\,y-\tfrac{y^{2}}{2}\Bigr]_{y=-1}^{y=\cos(x)} =\Bigl(\cos^{2}(x)-\tfrac{\cos^{2}(x)}{2}\Bigr) -\Bigl(-\cos(x)-\tfrac{1}{2}\Bigr) =\tfrac{\cos^{2}(x)}{2}+\cos(x)+\tfrac12.
  1. \displaystyle \int_{\cos(x)}^{1}\bigl(y-\cos(x)\bigr)\,dy
=\Bigl[\tfrac{y^{2}}{2}-\cos(x)\,y\Bigr]_{y=\cos(x)}^{y=1} =\Bigl(\tfrac12-\cos(x)\Bigr) -\Bigl(\tfrac{\cos^{2}(x)}{2}-\cos^{2}(x)\Bigr) =\tfrac12-\cos(x)+\tfrac{\cos^{2}(x)}{2}.
Somando as duas parcelas:
\int_{-1}^{1}\bigl|y-\cos(x)\bigr|\,dy =\Bigl(\tfrac{\cos^{2}x}{2}+\cos x+\tfrac12\Bigr) +\Bigl(\tfrac12-\cos x+\tfrac{\cos^{2}x}{2}\Bigr) =\cos^{2}(x)+1.

Passo 4

Agora a integral dupla vira
\int_{0}^{\pi}\bigl(1+\cos^{2}(x)\bigr)\,dx =\int_{0}^{\pi}1\,dx+\int_{0}^{\pi}\cos^{2}(x)\,dx =\pi+\int_{0}^{\pi}\cos^{2}(x)\,dx.
Usando a identidade \cos^{2}(x)=\tfrac{1+\cos(2x)}{2}, obtemos
\int_{0}^{\pi}\cos^{2}(x)\,dx =\int_{0}^{\pi}\frac{1+\cos(2x)}{2}\,dx =\frac{1}{2}\Bigl[\pi+0\Bigr] =\frac{\pi}{2}.

Resposta

\int_{0}^{\pi}\int_{-1}^{1}\bigl|y-\cos(x)\bigr|\,dy\,dx =\frac{3\pi}{2}.

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