Calcule:
\int_{0}^{\pi / 2}\left|\frac{1}{2}-\cos x\right| d x

Passo 1

O enunciado pede o valor de
\int_{0}^{\pi/2}\Bigl|\tfrac12-\cos x\Bigr|\,dx.
O conteúdo é integração de valor absoluto. Para resolver, vamos achar onde o integrando muda de sinal (isto é, onde \cos x=\tfrac12), dividir a integral em dois trechos e integrar sem o valor absoluto em cada trecho.

Passo 2

Resolvendo
\cos x = \tfrac12
em [0,\tfrac{\pi}{2}] obtemos x=\tfrac{\pi}{3}. Para 0\le x<\tfrac{\pi}{3}, \cos x>\tfrac12 e \tfrac12-\cos x<0, logo
\bigl|\tfrac12-\cos x\bigr| = \cos x - \tfrac12.
Para \tfrac{\pi}{3}<x\le\tfrac{\pi}{2}, \cos x<\tfrac12 e \tfrac12-\cos x>0, logo
\bigl|\tfrac12-\cos x\bigr| = \tfrac12 - \cos x.

Passo 3

Calcular
I_1 = \int_{0}^{\pi/3}\bigl(\cos x - \tfrac12\bigr)\,dx.
Temos
\int \cos x\,dx = \sin x, \qquad \int \tfrac12\,dx = \tfrac{x}{2},
logo
I_1 = \Bigl[\sin x - \tfrac{x}{2}\Bigr]_{0}^{\pi/3} = \sin\bigl(\tfrac{\pi}{3}\bigr) - \tfrac{\pi/6}{} = \tfrac{\sqrt{3}}{2} - \tfrac{\pi}{6}.

Passo 4

Calcular
I_2 = \int_{\pi/3}^{\pi/2}\bigl(\tfrac12 - \cos x\bigr)\,dx.
Temos
\int \bigl(\tfrac12 - \cos x\bigr)\,dx = \tfrac{x}{2} - \sin x,
logo
I_2 = \Bigl[\tfrac{x}{2} - \sin x\Bigr]_{\pi/3}^{\pi/2} = \Bigl(\tfrac{\pi}{4}-1\Bigr) - \Bigl(\tfrac{\pi}{6}-\tfrac{\sqrt{3}}{2}\Bigr) = \tfrac{\pi}{12} -1 + \tfrac{\sqrt{3}}{2}.

Passo 5

Somando I_1 e I_2:
I_1 + I_2 = \Bigl(\tfrac{\sqrt{3}}{2} - \tfrac{\pi}{6}\Bigr) + \Bigl(\tfrac{\pi}{12} -1 + \tfrac{\sqrt{3}}{2}\Bigr) = \sqrt{3} -1 - \tfrac{\pi}{12}.

Resposta

\sqrt{3} - 1 - \frac{\pi}{12}

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