let f(x) =∫_0 ^x (t/(1+sint))dt 1)find a explicit form of f(x) 2) calculate ∫


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Question Number 43337 by math khazana by abdo last updated on 09/Sep/18

$l e t f (x) = \int_{0}^{x} \frac{t}{1 + s i n t} d t$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{t}{1 + s i n t} d t$
Commented by maxmathsup by imad last updated on 11/Sep/18
$1) changement tan((t/2))=u give f(x)= ∫_0 ^(tan((x/2))) ((2 arctanu)/(1+ ((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^(tan((x/2))) ((4 arctanu)/(1+u^2 +2u))du =4∫_0 ^(tan((x/2))) (( arctanu)/((u+1)^2 )) du by parts we get f(x) =4{ [−(1/(1+u)) arctanu]_0 ^(tan((x/2))) +∫_0 ^(tan((x/2))) (du/((u+1)(1+u^2 ))} =−4 (x/(2(1+tan((x/2))))) +4 ∫_0 ^(tan((x/2))) (du/((u+1)(u^2 +1))) but ∫_0 ^(tan((x/2))) (du/((u+1)(u^2 +1))) =(1/2) ∫_0 ^(tan((x/2))) { (1/(u+1)) −((u−1)/(u^2 +1))}du =(1/2)[ln∣u+1∣]_0 ^(tan((x/2))) −(1/4) [ln(u^2 +1)]_0 ^(tan((x/2))) +(1/2) [ arctanu]_0 ^(tan((x/2))) =(1/2)ln∣1+tan((x/2))∣ −(1/4)ln∣1+tan^2 ((x/2))∣ +(x/4) ⇒ f(x) =((−2x)/(1+tan((x/2)))) +(1/2)ln∣1+tan((x/2))∣ −(1/4)ln∣1+tan^2 ((x/2))∣ +(x/4) .$
$1) c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e f (x) = \int_{0}^{t a n (\frac{x}{2})} \frac{2 a r c t a n u}{1 + \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= \int_{0}^{t a n (\frac{x}{2})} \frac{4 a r c t a n u}{1 + u^{2} + 2 u} d u = 4 \int_{0}^{t a n (\frac{x}{2})} \frac{a r c t a n u}{{(u + 1)}^{2}} d u b y p a r t s w e g e t$ $f (x) = 4 {{[- \frac{1}{1 + u} a r c t a n u]}_{0}^{t a n (\frac{x}{2})} + \int_{0}^{t a n (\frac{x}{2})} \frac{d u}{(u + 1) (1 + u^{2}}}$ $= - 4 \frac{x}{2 (1 + t a n (\frac{x}{2}))} + 4 \int_{0}^{t a n (\frac{x}{2})} \frac{d u}{(u + 1) (u^{2} + 1)} b u t$ $\int_{0}^{t a n (\frac{x}{2})} \frac{d u}{(u + 1) (u^{2} + 1)} = \frac{1}{2} \int_{0}^{t a n (\frac{x}{2})} {\frac{1}{u + 1} - \frac{u - 1}{u^{2} + 1}} d u$ $= \frac{1}{2} {[l n ∣ u + 1 ∣]}_{0}^{t a n (\frac{x}{2})} - \frac{1}{4} {[l n (u^{2} + 1)]}_{0}^{t a n (\frac{x}{2})} + \frac{1}{2} {[a r c t a n u]}_{0}^{t a n (\frac{x}{2})}$ $= \frac{1}{2} l n ∣ 1 + t a n (\frac{x}{2}) ∣ - \frac{1}{4} l n ∣ 1 + {t a n}^{2} (\frac{x}{2}) ∣ + \frac{x}{4} \Rightarrow$ $f (x) = \frac{- 2 x}{1 + t a n (\frac{x}{2})} + \frac{1}{2} l n ∣ 1 + t a n (\frac{x}{2}) ∣ - \frac{1}{4} l n ∣ 1 + {t a n}^{2} (\frac{x}{2}) ∣ + \frac{x}{4} .$
Commented by maxmathsup by imad last updated on 11/Sep/18

$2) t h e Q i s c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{t}{1 + s i n t} d t$
Commented by maxmathsup by imad last updated on 11/Sep/18

$\int_{0}^{\frac{π}{2}} \frac{t}{1 + s i n t} d t = f (\frac{π}{2}) = \frac{- π}{1 + 1} + \frac{1}{2} l n (2) - \frac{1}{4} l n (2) + \frac{π}{4}$ $= - \frac{π}{4} + \frac{l n (2)}{2} .$
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