find ∫_0 ^π (t/(2+sint)) dt


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Question Number 27600 by abdo imad last updated on 10/Jan/18

$f i n d \int_{0}^{π} \frac{t}{2 + s i n t} d t$
Commented by abdo imad last updated on 11/Jan/18
$let put I=∫_0 ^π (t/(2+sint))dt and the ch. tan((t/2))=x⇔ t=2arctanx I= ∫_0 ^∞ ((2arctanx)/(2+((2x)/(1+x^2 )))) ((2dx)/(1+x^2 ))= 2 ∫_0 ^∞ ((arctanx)/(1+x^2 +2x))dx =2∫_0 ^∞ ((arctanx)/((x+1)^2 ))dx then the integration per parts give ∫_0 ^∞ ((arctanx)/((x+1)^2 ))dx= [−(1/(x+1)) arctanx]_0 ^∝ − ∫_0 ^∞ −(1/(x+1)) (dx/(1+x^2 )) = ∫_0 ^∞ (dx/((x+1)(1+x^2 ))) let decompose rational fraction F(x)= (1/((x+1)(1+x^2 )))⇒F(x)= (a/(x+1)) + ((bx+c)/(1+x^2 )) a= lim_(x−>−1^ ) (x+1)F(x)= (1/2) lim_(x−>∝) xF(x)=0=a+b⇒b=−(1/2) so F(x)= (1/(2(x+1))) + ((((−x)/2)+c)/(1+x^2 )) ,F(0)=1= (1/2)+c⇒c=(1/2) F(x)= (1/(2(x+1))) −(1/2) ((x−1)/(1+x^2 )) I= ∫_0 ^∞ (dx/(x+1)) −∫_0 ^∞ ((x−1)/(1+x^2 ))dx = ∫_0 ^∞ (dx/(x+1)) −(1/2) ∫_0 ^∞ ((2x−2)/(1+x^2 ))dx I= ∫_0 ^∞ ( (1/(x+1)) −(1/2)((2x)/(1+x^2 )))dx + ∫_0 ^∞ (dx/(1+x^2 )) I=[ln/x+1/−(1/2)ln/1+x^2 /]_0 ^∝ + (π/2) I=[ln/((x+1)/(√(1+x^2 )))/ ]_0 ^∝ +(π/2)=0+(π/2) ∫_0 ^π (t/(2+sint))dt = (π/2) .$
$l e t p u t I = \int_{0}^{π} \frac{t}{2 + s i n t} d t a n d t h e c h . t a n (\frac{t}{2}) = x \Leftrightarrow$ $t = 2 a r c t a n x$ $I = \int_{0}^{\infty} \frac{2 a r c t a n x}{2 + \frac{2 x}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}} = 2 \int_{0}^{\infty} \frac{a r c t a n x}{1 + x^{2} + 2 x} d x$ $= 2 \int_{0}^{\infty} \frac{a r c t a n x}{{(x + 1)}^{2}} d x t h e n t h e i n t e g r a t i o n p e r p a r t s g i v e$ $\int_{0}^{\infty} \frac{a r c t a n x}{{(x + 1)}^{2}} d x = {[- \frac{1}{x + 1} a r c t a n x]}_{0}^{\propto} - \int_{0}^{\infty} - \frac{1}{x + 1} \frac{d x}{1 + x^{2}}$ $= \int_{0}^{\infty} \frac{d x}{(x + 1) (1 + x^{2})} l e t d e c o m p o s e r a t i o n a l f r a c t i o n$ $F (x) = \frac{1}{(x + 1) (1 + x^{2})} \Rightarrow F (x) = \frac{a}{x + 1} + \frac{b x + c}{1 + x^{2}}$ $a = {l i m}_{x - > - 1^{}} (x + 1) F (x) = \frac{1}{2}$ ${l i m}_{x - >\propto} x F (x) = 0 = a + b \Rightarrow b = - \frac{1}{2} s o$ $F (x) = \frac{1}{2 (x + 1)} + \frac{\frac{- x}{2} + c}{1 + x^{2}}, F (0) = 1 = \frac{1}{2} + c \Rightarrow c = \frac{1}{2}$ $F (x) = \frac{1}{2 (x + 1)} - \frac{1}{2} \frac{x - 1}{1 + x^{2}}$ $I = \int_{0}^{\infty} \frac{d x}{x + 1} - \int_{0}^{\infty} \frac{x - 1}{1 + x^{2}} d x$ $= \int_{0}^{\infty} \frac{d x}{x + 1} - \frac{1}{2} \int_{0}^{\infty} \frac{2 x - 2}{1 + x^{2}} d x$ $I = \int_{0}^{\infty} (\frac{1}{x + 1} - \frac{1}{2} \frac{2 x}{1 + x^{2}}) d x + \int_{0}^{\infty} \frac{d x}{1 + x^{2}}$ $I = {[l n / x + 1 / - \frac{1}{2} l n / 1 + x^{2} /]}_{0}^{\propto} + \frac{π}{2}$ $I = {[l n / \frac{x + 1}{\sqrt{1 + x^{2}}} /]}_{0}^{\propto} + \frac{π}{2} = 0 + \frac{π}{2}$ $\int_{0}^{π} \frac{t}{2 + s i n t} d t = \frac{π}{2} .$
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