Let f(t) = int left( frac{1 - sin(ln t)}{1 - | vedclass

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(A) Let $\ln t = x$,then $t = e^x$ and $dt = e^x dx$. Substituting these into the integral:$f(t) = \int \frac{1 - \sin x}{1 - \cos x} e^x dx = \int \f

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$-1 - \sqrt{2}$

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(A) Let $\ln t = x$,then $t = e^x$ and $dt = e^x dx$. Substituting these into the integral:$f(t) = \int \frac{1 - \sin x}{1 - \cos x} e^x dx = \int \frac{1 - 2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} e^x dx$$f(t) = \int \left( \frac{1}{2}\csc^2(x/2) - \cot(x/2) \right) e^x dx$Using the identity $\int e^x (g(x) + g'(x)) dx = e^x g(x) + C$,where $g(x) = -\cot(x/2)$ and $g'(x) = \frac{1}{2}\csc^2(x/2)$:$f(t) = -e^x \cot(x/2) + C = -t \cot(\frac{\ln t}{2}) + C$Given $f(e^{\pi/2}) = -e^{\pi/2} \cot(\pi/4) + C = -e^{\pi/2} + C = -e^{\pi/2}$,so $C = 0$.Thus,$f(t) = -t \cot(\frac{\ln t}{2})$.$f(e^{\pi/4}) = -e^{\pi/4} \cot(\pi/8) = -e^{\pi/4} (\sqrt{2} + 1)$.Comparing with $f(e^{\pi/4}) = \alpha e^{\pi/4}$,we get $\alpha = -(1 + \sqrt{2}) = -1 - \sqrt{2}$.

Let $f(t) = \int \left( \frac{1 - \sin(\ln t)}{1 - \cos(\ln t)} \right) dt$,for $t > 1$. If $f(e^{\pi/2}) = -e^{\pi/2}$ and $f(e^{\pi/4}) = \alpha e^{\pi/4}$,then $\alpha$ equals:

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