int_{pi /2}^{pi} frac{1 - sin x}{1 - cos x} , | vedclass

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(A) Let $I = \int_{\pi /2}^{\pi} \frac{1 - \sin x}{1 - \cos x} \, dx$. Using the identities $1 - \cos x = 2 \sin^2(x/2)$ and $\sin x = 2 \sin(x/2) \co

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$1 - \ln 2$

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(A) Let $I = \int_{\pi /2}^{\pi} \frac{1 - \sin x}{1 - \cos x} \, dx$. Using the identities $1 - \cos x = 2 \sin^2(x/2)$ and $\sin x = 2 \sin(x/2) \cos(x/2)$,we have: $I = \int_{\pi /2}^{\pi} \frac{1 - 2 \sin(x/2) \cos(x/2)}{2 \sin^2(x/2)} \, dx$ $I = \int_{\pi /2}^{\pi} \left( \frac{1}{2} \csc^2(x/2) - \frac{2 \sin(x/2) \cos(x/2)}{2 \sin^2(x/2)} \right) \, dx$ $I = \int_{\pi /2}^{\pi} \left( \frac{1}{2} \csc^2(x/2) - \cot(x/2) \right) \, dx$ Integrating term by term: $\int \frac{1}{2} \csc^2(x/2) \, dx = \frac{1}{2} \cdot (-2 \cot(x/2)) = -\cot(x/2)$ $\int -\cot(x/2) \, dx = -2 \ln|\sin(x/2)|$ So,$I = [-\cot(x/2) - 2 \ln|\sin(x/2)|]_{\pi /2}^{\pi}$ Evaluating at the limits: At $x = \pi$: $-\cot(\pi/2) - 2 \ln(\sin(\pi/2)) = 0 - 2 \ln(1) = 0$ At $x = \pi/2$: $-\cot(\pi/4) - 2 \ln(\sin(\pi/4)) = -1 - 2 \ln(1/\sqrt{2}) = -1 - 2 \ln(2^{-1/2}) = -1 - 2(-1/2) \ln 2 = -1 + \ln 2$ $I = 0 - (-1 + \ln 2) = 1 - \ln 2$.

$\int_{\pi /2}^{\pi} \frac{1 - \sin x}{1 - \cos x} \, dx =$

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