Question
Let $f(t) = \int\left( \frac{1 - sin\left( \log_{e}t \right)}{1 - cos\left( \log_{e}t \right)} \right)dt,t > 1$.
If $f\left( e^{\pi/2} \right) = - e^{\pi/2}$ and $f\left( e^{\pi/4} \right) = \alpha e^{\pi/4}$, then $\alpha$ equals
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Solution
$f(t) = \int\frac{1 - sin(lnt)}{1 - cos(lnt)}dt$
Let $lnt = x \Rightarrow t = e^{x} \Rightarrow dt = e^{x}dx$
$${= \frac{1}{2}\int\left( {cosec}^{2}\frac{x}{2} - 2cot\frac{x}{2} \right)e^{x}dx - tcot\left( \frac{lnt}{2} \right) + C }{\left( \therefore\int\left( f(x) + f'(x) \right)e^{x}dx = f(x) \cdot e^{x} + c \right) }$$Now $f\left( e^{\pi/2} \right) = - e^{\pi/2}cot\left( \frac{\pi}{4} \right) + C = - e^{\frac{\pi}{2}}$ (given)
$$C = 0 $$Now $f\left( e^{\pi/4} \right) = - e^{\pi/4}cot\left( \frac{\pi}{8} \right) + C = - e^{\frac{\pi}{4}}(\sqrt{2} + 1)$
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