calculate lim_(x→1) ∫_x ^x^2 ((arctan(2t))/(sin(πt)))dt


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Question Number 39839 by math khazana by abdo last updated on 12/Jul/18

$c a l c u l a t e {l i m}_{x \to 1} \int_{x}^{x^{2}} \frac{a r c t a n (2 t)}{s i n (π t)} d t$
Commented by math khazana by abdo last updated on 30/Jul/18

$l e t f (x) = \int_{x}^{x^{2}} \frac{a r c t a n (2 t)}{s i n (π t)} d t c h a n g e m e n t t = u + 1$ $g i v e f (x) = \int_{x - 1}^{x^{2} - 1} \frac{a r c t a n (2 u + 2)}{s i n (π u + π)} d u$ $= - \int_{x - 1}^{x^{2} - 1} \frac{a r c t a n (2 u + 2)}{s i n (π u)} d u b u t \exists c \in] x - 1, x^{2} - 1 [/$ $f (x) = - a r c t a n (2 c + 2) \int_{x - 1}^{x^{2} - 1} \frac{d u}{s i n (π u)} w e h a v e$ $\int_{x - 1}^{x^{2} - 1} \frac{d u}{s i n (π u)} =_{π u = t} \int_{π (x - 1)}^{π (x^{2} - 1)} \frac{d t}{π s i n t}$ $=_{t a n (\frac{t}{2}) = α} \frac{1}{π} \int_{t a n (\frac{π}{2} (x - 1))}^{t a n (\frac{π}{2} (x^{2} - 1))} \frac{1}{\frac{2 α}{1 + α^{2}}} \frac{2 d α}{1 + α^{2}}$ $= \frac{1}{π} {[l n ∣ α ∣]}_{t a n (\frac{π}{2} (x - 1))}^{t a n (\frac{π}{2} (x^{2} - 1))} = \frac{1}{π} l n ∣ \frac{t a n (\frac{π}{2} (x^{2} - 1))}{t a n (\frac{π}{2} (x - 1))} ∣$ $f o r x \in v (1) t a n (\frac{π}{2} (x^{2} - 1)) \sim \frac{π}{2} (x^{2} - 1) a n d$ $t a n (\frac{π}{2} (x - 1)) \sim \frac{π}{2} (x - 1) \Rightarrow∣ \frac{t a n (. . .)}{t a n (. .)} ∣\sim∣ x + 1 ∣\Rightarrow$ ${l i m}_{x \to 1} \int_{x - 1}^{x^{2} - 1} \frac{d u}{s i n (π u)} = \frac{l n (2)}{π} \Rightarrow$ ${l i m}_{x \to 1} f (x) = - a r c t a n (2) \frac{l n (2)}{π} .$
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