calculate ∫_0 ^π ((1+2sinx)/(3 +2cosx))dx let A =∫_0 ^π ((1+2sinx)/(3 +2cosx))dx changement tan((x/2))=t give A =∫_0 ^∞ ((1+((4t)/(1+t^2 )))/(3+2((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =2 ∫_0 ^∞ ((1+t^2 +4t)/((1+t^2 )^2 (((3+3t^2 +2−2t^2 )/(1+t^2 )))))dt =2 ∫_0 ^∞ ((t^2 +4t +1)/((1+t^2 )(5+t^2 )))dt let decompose F(t)=((t^2 +4t+1)/((t^2 +1)(t^2 +5))) F(t)=((at +b)/(t^2 +1)) +((ct +d)/(t^2 +5)) ⇒(at+b)(t^2 +5)+(ct+d)(t^2 +1) =t^2 +4t +1 ⇒ at^3 +5at +bt^2 +5b +ct^3 +ct +dt^2 +d =t^2 +4t +1 ⇒ (a+c)t^3 +(b+d)t^2 +(5a+c)t +5b +d =t^2 +4t +1 ⇒a+c=0 and b+d=1 and 5a+c =4 and 5b+d =1 ⇒c=−a ⇒a=1 ⇒c=−1 we have d=1−b ⇒5b +1−b =1 ⇒b=0 ⇒d=1 ⇒ F(t)=(t/(t^2 +1)) +((−t +1)/(t^2 +5)) ⇒ A =2 ∫_0 ^∞ F(t)dt =∫_0 ^∞ ((2t)/(t^2 +1))dt +∫_0 ^∞ ((−2t +2)/(t^2 +5))dt =[ln(((t^2 +1)/(t^2 +5)))]_0 ^(+∞) +2 ∫_0 ^∞ (dt/(t^2 +5)) =ln(5) + 2 ∫_0 ^∞ (dt/(t^2 +5)) but ∫_0 ^∞ (dt/(t^2 +5))dt =_(t =(√5)u ) ∫_0 ^∞ (((√5)du)/(5(1+u^2 ))) =(1/(√5)) [artanu]_0 ^(+∞) =(π/(2(√5))) ⇒ A =ln(5) +(π/(2(√5))) .


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Question Number 53112 by maxmathsup by imad last updated on 21/Jan/19

$c a l c u l a t e \int_{0}^{π} \frac{1 + 2 s i n x}{3 + 2 c o s x} d x$ $l e t A = \int_{0}^{π} \frac{1 + 2 s i n x}{3 + 2 c o s x} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $A = \int_{0}^{\infty} \frac{1 + \frac{4 t}{1 + t^{2}}}{3 + 2 \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 2 \int_{0}^{\infty} \frac{1 + t^{2} + 4 t}{{(1 + t^{2})}^{2} (\frac{3 + 3 t^{2} + 2 - 2 t^{2}}{1 + t^{2}})} d t$ $= 2 \int_{0}^{\infty} \frac{t^{2} + 4 t + 1}{(1 + t^{2}) (5 + t^{2})} d t l e t d e c o m p o s e F (t) = \frac{t^{2} + 4 t + 1}{(t^{2} + 1) (t^{2} + 5)}$ $F (t) = \frac{a t + b}{t^{2} + 1} + \frac{c t + d}{t^{2} + 5} \Rightarrow (a t + b) (t^{2} + 5) + (c t + d) (t^{2} + 1) = t^{2} + 4 t + 1 \Rightarrow$ ${a t}^{3} + 5 a t + {b t}^{2} + 5 b + {c t}^{3} + c t + {d t}^{2} + d = t^{2} + 4 t + 1 \Rightarrow$ $(a + c) t^{3} + (b + d) t^{2} + (5 a + c) t + 5 b + d = t^{2} + 4 t + 1 \Rightarrow a + c = 0 a n d b + d = 1 a n d$ $5 a + c = 4 a n d 5 b + d = 1 \Rightarrow c = - a \Rightarrow a = 1 \Rightarrow c = - 1$ $w e h a v e d = 1 - b \Rightarrow 5 b + 1 - b = 1 \Rightarrow b = 0 \Rightarrow d = 1 \Rightarrow$ $F (t) = \frac{t}{t^{2} + 1} + \frac{- t + 1}{t^{2} + 5} \Rightarrow A = 2 \int_{0}^{\infty} F (t) d t = \int_{0}^{\infty} \frac{2 t}{t^{2} + 1} d t + \int_{0}^{\infty} \frac{- 2 t + 2}{t^{2} + 5} d t$ $= {[l n (\frac{t^{2} + 1}{t^{2} + 5})]}_{0}^{+ \infty} + 2 \int_{0}^{\infty} \frac{d t}{t^{2} + 5} = l n (5) + 2 \int_{0}^{\infty} \frac{d t}{t^{2} + 5} b u t$ $\int_{0}^{\infty} \frac{d t}{t^{2} + 5} d t =_{t = \sqrt{5} u} \int_{0}^{\infty} \frac{\sqrt{5} d u}{5 (1 + u^{2})} = \frac{1}{\sqrt{5}} {[a r t a n u]}_{0}^{+ \infty} = \frac{π}{2 \sqrt{5}} \Rightarrow$ $A = l n (5) + \frac{π}{2 \sqrt{5}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
	$∫(1/(3+2cosx))dx+∫((2sinx)/(3+2cosx))dx ∫((1+tan^2 (x/2))/(3+3tan^2 (x/2)+2−2tan^2 (x/2)))dx−∫((d(3+2cosx))/(3+2cosx)) ∫((sec^2 (x/2))/(5+tan^2 (x/2)))dx−ln(3+2cosx)+c 2∫((d(tan(x/2)))/(5+tan^2 (x/2)))−do ∣2×(1/(√5))tan^(−1) (((tan(x/2))/(√5)))−ln(3+2cosx)∣_0 ^π [{(2/(√5))tan^(−1) (((tan(π/2))/(√5)))−ln(3+2cosπ)}−{(2/(√5))tan^(−1) (((tan0)/(√5)))−ln(3+2cos0)}] =[{(2/(√5))((π/2))−ln(3−2)}−{(2/(√5))tan^(−1) (0)−ln5}] =(π/(√5))+ln5$
	$\int \frac{1}{3 + 2 c o s x} d x + \int \frac{2 s i n x}{3 + 2 c o s x} d x$ $\int \frac{1 + {t a n}^{2} \frac{x}{2}}{3 + 3 {t a n}^{2} \frac{x}{2} + 2 - 2 {t a n}^{2} \frac{x}{2}} d x - \int \frac{d (3 + 2 c o s x)}{3 + 2 c o s x}$ $\int \frac{{s e c}^{2} \frac{x}{2}}{5 + {t a n}^{2} \frac{x}{2}} d x - l n (3 + 2 c o s x) + c$ $2 \int \frac{d (t a n \frac{x}{2})}{5 + {t a n}^{2} \frac{x}{2}} - d o$ $∣ 2 \times \frac{1}{\sqrt{5}} {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{\sqrt{5}}) - l n (3 + 2 c o s x) ∣_{0}^{π}$ $[{\frac{2}{\sqrt{5}} {t a n}^{- 1} (\frac{t a n \frac{π}{2}}{\sqrt{5}}) - l n (3 + 2 c o s π)} - {\frac{2}{\sqrt{5}} {t a n}^{- 1} (\frac{t a n 0}{\sqrt{5}}) - l n (3 + 2 c o s 0)}]$ $= [{\frac{2}{\sqrt{5}} (\frac{π}{2}) - l n (3 - 2)} - {\frac{2}{\sqrt{5}} {t a n}^{- 1} (0) - l n 5}]$ $= \frac{π}{\sqrt{5}} + l n 5$
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