Let f ( t ) = ( 1 - sin ( log e t ) 1 - cos ( log e t ) ) d t , t 1 . If f ( e / 2 ) = - e / 2 and f ( e / 4 ) = e / 4 , then equals [2026]

Question

Let $f (t)$ $= \int (\frac{1 - \sin (\log_{e} t)}{1 - \cos (\log_{e} t)}) d t, t > 1 .$

If $f (e^{π / 2}) = - e^{π / 2} and f (e^{π / 4}) = α e^{π / 4},$ then $α$ equals [2026]

Smart Achievers · Accepted Answer

(1)

$f (t) = \int \frac{1 - \sin (\ln t)}{1 - \cos (\ln t)} d t$

$Let \ln t = x \Rightarrow t = e^{x} \Rightarrow d t = e^{x} d x$

$= \frac{1}{2} \int (\cos e c^{2} \frac{x}{2} - 2 \cot \frac{x}{2}) e^{x} d x - t \cot (\frac{\ln t}{2}) + C$

$(∴ \int (f (x) + f' (x)) e^{x} d x = f (x) . e^{x} + C)$

$Now f (e^{π / 2}) = - e^{π / 2} \cot (\frac{π}{4}) + C = - e^{π / 2} (given)$

$\Rightarrow C = 0$

$Now f (e^{π / 4}) = - e^{π / 4} \cot (\frac{π}{8}) + C = - e^{π / 4} (\sqrt{2} + 1)$

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