find ∫_0 ^π ((xdx)/(1+sinx))


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Question Number 35242 by abdo mathsup 649 cc last updated on 17/May/18

$f i n d \int_{0}^{π} \frac{x d x}{1 + s i n x}$
Commented by abdo mathsup 649 cc last updated on 17/May/18
$let put I = ∫_0 ^π ((xdx)/(1+sinx)) .changement tan((x/2))=t give I = ∫_0 ^∞ ((2arctant)/(1+ ((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = 4 ∫_0 ^∞ ((arctan(t))/(1+t^2 +2t))dt = 4 ∫_0 ^∞ ((arctan(t))/((t+1)^2 ))dt and by parts ∫_0 ^∞ ((arctan(t))/((t+1)^2 ))dt = [((−1)/(t+1)) arctant]_0 ^(+∞) −∫_0 ^∞ ((−1)/((t+1)(1+t^2 )))dt = ∫_0 ^∞ (dt/((t+1)(t^2 +1))) =∫_0 ^∞ { (1/(t+1)) −((t−1)/(t^2 +1))}dt =∫_0 ^∞ { (1/(t+1)) −(1/2) ((2t)/(t^2 +1))}dt +∫_0 ^∞ (dt/(t^2 +1)) =[ln(((t+1)/(t^2 +1)))]_0 ^(+∞) +(π/2) =(π/2) ⇒ I = 4.(π/2) ⇒ I =2π .$
$l e t p u t I = \int_{0}^{π} \frac{x d x}{1 + s i n x} . c h a n g e m e n t t a n (\frac{x}{2}) = t$ $g i v e I = \int_{0}^{\infty} \frac{2 a r c t a n t}{1 + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= 4 \int_{0}^{\infty} \frac{a r c t a n (t)}{1 + t^{2} + 2 t} d t = 4 \int_{0}^{\infty} \frac{a r c t a n (t)}{{(t + 1)}^{2}} d t$ $a n d b y p a r t s$ $\int_{0}^{\infty} \frac{a r c t a n (t)}{{(t + 1)}^{2}} d t = {[\frac{- 1}{t + 1} a r c t a n t]}_{0}^{+ \infty} - \int_{0}^{\infty} \frac{- 1}{(t + 1) (1 + t^{2})} d t$ $= \int_{0}^{\infty} \frac{d t}{(t + 1) (t^{2} + 1)}$ $= \int_{0}^{\infty} {\frac{1}{t + 1} - \frac{t - 1}{t^{2} + 1}} d t$ $= \int_{0}^{\infty} {\frac{1}{t + 1} - \frac{1}{2} \frac{2 t}{t^{2} + 1}} d t + \int_{0}^{\infty} \frac{d t}{t^{2} + 1}$ $= {[l n (\frac{t + 1}{t^{2} + 1})]}_{0}^{+ \infty} + \frac{π}{2} = \frac{π}{2} \Rightarrow I = 4 . \frac{π}{2} \Rightarrow$ $I = 2 π .$
	Answered by math1967 last updated on 17/May/18
	$2I=∫_(0 ) ^π ((xdx)/(1+sinx)) +∫_0 ^π (((π−x)dx)/(1+sin(π−x))) =π∫_0 ^π (dx/(1+sinx))=π∫_0 ^π (dx/(1+cos((π/2)−x))) =π∫_0 ^π (dx/(2cos^2 ((π/4)−(x/2)))) [1+cosx=2cos^2 (x/2)] =π∫_0 ^π sec^2 ((π/4)−(x/2))dx =(π/2)∫_(0 ) ^π sec^2 ((π/4)−(x/2))dx=(π/2)[((tan((π/4)−(x/2)))/((−1)/2))]_0 ^π =−π{tan(−(π/4))−tan(π/4)}=2π ∴I=π$
	$2 I = \underset{0}{\int^{π}} \frac{x d x}{1 + s i n x} + \underset{0}{\int^{π}} \frac{(π - x) d x}{1 + s i n (π - x)}$ $= π \underset{0}{\int^{π}} \frac{d x}{1 + s i n x} = π \underset{0}{\int^{π}} \frac{d x}{1 + c o s (\frac{π}{2} - x)}$ $= π \underset{0}{\int^{π}} \frac{d x}{2 {c o s}^{2} (\frac{π}{4} - \frac{x}{2})} [1 + c o s x = 2 {c o s}^{2} \frac{x}{2}]$ $= π \underset{0}{\int^{π}} {s e c}^{2} (\frac{π}{4} - \frac{x}{2}) d x$ $= \frac{π}{2} \underset{0}{\int^{π}} {s e c}^{2} (\frac{π}{4} - \frac{x}{2}) d x = \frac{π}{2} {[\frac{t a n (\frac{π}{4} - \frac{x}{2})}{\frac{- 1}{2}}]}_{0}^{π}$ $= - π {t a n (- \frac{π}{4}) - t a n \frac{π}{4}} = 2 π$ $∴ I = π$
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