((∡ BeMath ∡)/▽) I = ∫_0 ^π ((x dx)/(1+sin x))


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Question Number 108391 by bemath last updated on 16/Aug/20

$\frac{∡ B e M a t h ∡}{▽}$ $I = \underset{0}{\int^{π}} \frac{x d x}{1 + \sin x}$
	Answered by bobhans last updated on 16/Aug/20

	Commented by khanshadab2405 last updated on 16/Aug/20
	1/√1²+cot²∅
	Commented by bobhans last updated on 16/Aug/20

	$t y p o I = \frac{1}{2} \underset{- π / 4}{\int^{π / 4}} (4 w + π) \sec^{2} w d w$ $I = \frac{1}{2} \times 2 π = π$
	Answered by Dwaipayan Shikari last updated on 16/Aug/20

	$\int_{0}^{π} \frac{x d x}{1 + s i n x} = \int_{0}^{π} \frac{π - x}{1 + s i n x} d x = I$ $2 I = \int_{0}^{π} \frac{π}{1 + s i n x} d x$ $2 I = 2 π \int_{0}^{\infty} \frac{1}{1 + \frac{2 t}{1 + t^{2}}} . \frac{1}{1 + t^{2}} d t (t a n \frac{x}{2} = t$ $2 I = 2 π \int_{0}^{\infty} \frac{1}{{(1 + t)}^{2}}$ $I = π {[- \frac{1}{1 + t}]}_{0}^{\infty}$ $I = π$
	Answered by mnjuly1970 last updated on 16/Aug/20

	$I = \int_{0}^{π} \frac{(π - x) d x}{1 + s i n (π - x)} d x$ $\Rightarrow 2 I = π \int_{0}^{π} \frac{1 - s i n (x)}{{c o s}^{2} (x)} d x$ $2 I = π ({[t a n (x)]}_{0}^{π} - {[\frac{1}{c o s (x)}]}_{0}^{π}) = π (2)$ $I = π ♣ . . . . . . . . . . . ♣$
	Answered by mathmax by abdo last updated on 16/Aug/20

	$I = \int_{0}^{π} \frac{xdx}{1 + sinx} changement x = π - t give$ $I = \int_{0}^{π} \frac{(π - t)}{1 + sint} dt = π \int_{0}^{π} \frac{dt}{1 + sint} - I \Rightarrow 2 I = π \int_{0}^{π} \frac{dt}{1 + sint}$ $=_{\tan (\frac{t}{2}) = u} π \int_{0}^{\infty} \frac{2 du}{(1 + u^{2}) (1 + \frac{2 u}{1 + u^{2}})}$ $= 2 π \int_{0}^{\infty} \frac{du}{1 + u^{2} + 2 u} = 2 π \int_{0}^{\infty} \frac{du}{{(u + 1)}^{2}} = 2 π {[- \frac{1}{u + 1}]}_{0}^{+ \infty} = 2 π \Rightarrow$ $2 I = 2 π \Rightarrow I = π$
	Answered by mathmax by abdo last updated on 16/Aug/20
	$another way changement tan((x/2))=t give I =∫_0 ^∞ ((2arctan(t))/((1+((2t)/(1+t^2 )))))((2dt)/(1+t^2 )) =4∫_0 ^∞ ((arctant)/(1+t^2 +2t))dt =4∫_0 ^∞ ((arctant)/((t+1)^2 ))dt =_(byparts) 4{[−((arctant)/(t+1))]_0 ^(+∞) +∫_0 ^∞ (1/(t+1))(dt/(1+t^2 ))} =4 ∫_0 ^∞ (dt/((t+1)(t^2 +1))) let decompose F(t) =(1/((t+1)(t^2 +1))) F(t) =(a/(t+1)) +((bt +c)/(t^2 +1)) a =(1/2) , lim_(t→+∞) tF(t) =0 =a+b ⇒b=−(1/2) F(0) =1 =a +c ⇒c=1−a =(1/2) ⇒F(t)=(1/(2(t+1)))+((−(1/2)t +(1/2))/(t^2 +1)) ⇒ ∫_0 ^∞ F(t)dt =(1/2)∫_0 ^∞ (dt/(t+1))−(1/4)∫_0 ^∞ ((2t)/(t^2 +1)) +(1/2)∫_0 ^∞ (dt/(1+t^2 )) =[(1/2)ln(t+1)−(1/4)ln(t^2 +1)]_0 ^∞ +(1/2)[arctant]_0 ^∞ =[ln(((√(t+1))/((^4 (√(t^2 +1))))))]_0 ^(+∞) +(1/2){(π/2)} =(π/4) ⇒I =4.(π/4) ⇒I =π$
	$another way changement \tan (\frac{x}{2}) = t give$ $I = \int_{0}^{\infty} \frac{2 \arctan (t)}{(1 + \frac{2 t}{1 + t^{2}})} \frac{2 dt}{1 + t^{2}} = 4 \int_{0}^{\infty} \frac{arctant}{1 + t^{2} + 2 t} dt$ $= 4 \int_{0}^{\infty} \frac{arctant}{{(t + 1)}^{2}} dt =_{byparts} 4 {{[- \frac{arctant}{t + 1}]}_{0}^{+ \infty} + \int_{0}^{\infty} \frac{1}{t + 1} \frac{dt}{1 + t^{2}}}$ $= 4 \int_{0}^{\infty} \frac{dt}{(t + 1) (t^{2} + 1)} let decompose F (t) = \frac{1}{(t + 1) (t^{2} + 1)}$ $F (t) = \frac{a}{t + 1} + \frac{bt + c}{t^{2} + 1}$ $a = \frac{1}{2}, \lim_{t \to + \infty} tF (t) = 0 = a + b \Rightarrow b = - \frac{1}{2}$ $F (0) = 1 = a + c \Rightarrow c = 1 - a = \frac{1}{2} \Rightarrow F (t) = \frac{1}{2 (t + 1)} + \frac{- \frac{1}{2} t + \frac{1}{2}}{t^{2} + 1} \Rightarrow$ $\int_{0}^{\infty} F (t) dt = \frac{1}{2} \int_{0}^{\infty} \frac{dt}{t + 1} - \frac{1}{4} \int_{0}^{\infty} \frac{2 t}{t^{2} + 1} + \frac{1}{2} \int_{0}^{\infty} \frac{dt}{1 + t^{2}}$ $= {[\frac{1}{2} \ln (t + 1) - \frac{1}{4} \ln (t^{2} + 1)]}_{0}^{\infty} + \frac{1}{2} {[arctant]}_{0}^{\infty}$ $= {[\ln (\frac{\sqrt{t + 1}}{(^{4} \sqrt{t^{2} + 1})})]}_{0}^{+ \infty} + \frac{1}{2} {\frac{π}{2}} = \frac{π}{4} \Rightarrow I = 4 . \frac{π}{4} \Rightarrow I = π$
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