Calcule as seguintes integrais definidas:
\int_{e^{-2 \pi}}^{1}\left|\cos \left(\ln \frac{1}{x}\right)\right| d x.

Passo 1

O enunciado pede para calcular
\int_{e^{-2\pi}}^{1}\bigl|\cos\bigl(\ln\tfrac{1}{x}\bigr)\bigr|\,dx.
Vamos usar mudança de variável para eliminar o logaritmo e depois tratar o valor absoluto de \cos por partes, aproveitando que a função \cos t tem período 2\pi e muda de sinal em t=\tfrac\pi2,\tfrac{3\pi}2.

Passo 2

Fazemos a substituição
t = \ln\frac1x = -\ln x,
então x=e^{-t} e
dx = -e^{-t}\,dt.
Quando x=e^{-2\pi}, t=-\ln(e^{-2\pi})=2\pi. Quando x=1, t=0. Logo
\int_{e^{-2\pi}}^{1}\bigl|\cos(\ln\tfrac1x)\bigr|\,dx =\int_{t=2\pi}^{0}|\cos t|\bigl(-e^{-t}\bigr)\,dt =\int_{0}^{2\pi}e^{-t}|\cos t|\,dt.

Passo 3

Agora escrevemos
\int_{0}^{2\pi}e^{-t}|\cos t|\,dt =\int_{0}^{\frac\pi2}e^{-t}\cos t\,dt +\int_{\frac\pi2}^{\frac{3\pi}2}e^{-t}\bigl(-\cos t\bigr)\,dt +\int_{\frac{3\pi}2}^{2\pi}e^{-t}\cos t\,dt,
pois em [0,\pi/2] e [3\pi/2,2\pi] temos \cos t\ge0, e em [\pi/2,3\pi/2] temos \cos t\le0.

Passo 4

Calculamos a primitiva de e^{-t}\cos t. Sabemos que
\frac{d}{dt}\Bigl[e^{-t}\,\frac{\sin t-\cos t}{2}\Bigr] =e^{-t}\cos t.
Portanto
\int e^{-t}\cos t\,dt =e^{-t}\,\frac{\sin t-\cos t}{2}+C.
E
\int e^{-t}\bigl(-\cos t\bigr)\,dt =-\,e^{-t}\,\frac{\sin t-\cos t}{2}+C =e^{-t}\,\frac{\cos t-\sin t}{2}+C.

Passo 5

Cálculo dos três termos:
  1. \displaystyle \int_{0}^{\frac\pi2}e^{-t}\cos t\,dt
\Bigl[\;e^{-t}\,\tfrac{\sin t-\cos t}{2}\Bigr]_{0}^{\frac\pi2} =\frac{e^{-\pi/2}(1-0)}{2}-\frac{1\cdot(0-1)}{2} =\frac{e^{-\pi/2}+1}{2}.
  1. \displaystyle \int_{\frac\pi2}^{\frac{3\pi}2}e^{-t}(-\cos t)\,dt
\Bigl[\;e^{-t}\,\tfrac{\cos t-\sin t}{2}\Bigr]_{\pi/2}^{3\pi/2} =\frac{e^{-3\pi/2}(0-(-1))}{2}-\frac{e^{-\pi/2}(0-1)}{2} =\frac{e^{-3\pi/2}+e^{-\pi/2}}{2}.
  1. \displaystyle \int_{\frac{3\pi}2}^{2\pi}e^{-t}\cos t\,dt
\Bigl[\;e^{-t}\,\tfrac{\sin t-\cos t}{2}\Bigr]_{3\pi/2}^{2\pi} =\frac{e^{-2\pi}(0-1)}{2}-\frac{e^{-3\pi/2}(-1-0)}{2} =\frac{-e^{-2\pi}+e^{-3\pi/2}}{2}.

Passo 6

Somando os três resultados:
\frac{e^{-\pi/2}+1}{2} +\frac{e^{-3\pi/2}+e^{-\pi/2}}{2} +\frac{-e^{-2\pi}+e^{-3\pi/2}}{2} =\frac{1 + 2e^{-\pi/2} + 2e^{-3\pi/2} - e^{-2\pi}}{2}.

Resposta

\int_{e^{-2\pi}}^{1}\bigl|\cos(\ln\tfrac{1}{x})\bigr|\,dx =\frac{1 + 2e^{-\frac\pi2} + 2e^{-\frac{3\pi}2} - e^{-2\pi}}{2}.

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