find the value of ∫_0 ^(2π) (dx/(3+2sinx +cosx))


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Question Number 67526 by mathmax by abdo last updated on 28/Aug/19

$f i n d t h e v a l u e o f \int_{0}^{2 π} \frac{d x}{3 + 2 s i n x + c o s x}$
Commented by mathmax by abdo last updated on 31/Aug/19
$let I =∫_0 ^(2π) (dx/(3+2sinx +cosx)) ⇒I =∫_0 ^π (dx/(3+2sinx +cosx)) +∫_π ^(2π) (dx/(3+2sinx +cosx)) =H +K H =_(tan((x/2))=t) ∫_0 ^∞ (1/(3+2((2t)/(1+t^2 )) +((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^∞ ((2dt)/(3+3t^2 +4t +1−t^2 )) =∫_0 ^∞ ((2dt)/(2t^2 +4t+4)) =∫_0 ^∞ (dt/(t^2 +2t +2)) Δ^′ =1−2 =−1<0 ⇒ H =∫_0 ^∞ (dt/(t^2 +2t+1 +1)) =∫_0 ^∞ (dt/((t+1)^2 +1)) =_(t+1 =α) ∫_1 ^(+∞) (dα/(1+α^2 )) =[arctan(α)]_1 ^(+∞) =(π/2)−(π/4) =(π/4) K =_(x =π +t) ∫_0 ^π (dt/(3−2sint−cost)) =_(tan((t/2))=u) ∫_0 ^∞ ((2du)/((1+u^2 )(3−2((2u)/(1+u^2 ))−((1−u^2 )/(1+u^2 ))))) =∫_0 ^∞ ((2du)/(3+3u^2 −4u−1+u^2 )) =∫_0 ^∞ ((2du)/(4u^2 −4u+2)) =∫_0 ^∞ (du/(2u^2 −2u +1)) =(1/2)∫_0 ^∞ (du/(u^2 −u +(1/2))) =(1/2)∫_0 ^∞ (du/((u−(1/2))^2 +(3/4)+(1/2))) =(1/2)∫_0 ^∞ (du/((u−(1/2))^2 +(5/4))) =_(u−(1/2)=((√5)/2)α) (1/2)×(4/5)∫_(−(1/(√5))) ^(+∞) (1/(1+u^2 ))((√5)/2)dα =(1/(√5))[arctanα]_(−(1/(√5))) ^(+∞) =(1/(√5)){(π/2) +arctan((1/(√5)))} ⇒ I =(π/4) +(π/(2(√5))) +(1/(√5)) arctan((1/(√5))).$
$l e t I = \int_{0}^{2 π} \frac{d x}{3 + 2 s i n x + c o s x} \Rightarrow I = \int_{0}^{π} \frac{d x}{3 + 2 s i n x + c o s x} + \int_{π}^{2 π} \frac{d x}{3 + 2 s i n x + c o s x}$ $= H + K$ $H =_{t a n (\frac{x}{2}) = t} \int_{0}^{\infty} \frac{1}{3 + 2 \frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int_{0}^{\infty} \frac{2 d t}{3 + 3 t^{2} + 4 t + 1 - t^{2}} = \int_{0}^{\infty} \frac{2 d t}{2 t^{2} + 4 t + 4} = \int_{0}^{\infty} \frac{d t}{t^{2} + 2 t + 2}$ $Δ^{^{'}} = 1 - 2 = - 1 < 0 \Rightarrow$ $H = \int_{0}^{\infty} \frac{d t}{t^{2} + 2 t + 1 + 1} = \int_{0}^{\infty} \frac{d t}{{(t + 1)}^{2} + 1} =_{t + 1 = α} \int_{1}^{+ \infty} \frac{d α}{1 + α^{2}}$ $= {[a r c t a n (α)]}_{1}^{+ \infty} = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}$ $K =_{x = π + t} \int_{0}^{π} \frac{d t}{3 - 2 s i n t - c o s t} =_{t a n (\frac{t}{2}) = u} \int_{0}^{\infty} \frac{2 d u}{(1 + u^{2}) (3 - 2 \frac{2 u}{1 + u^{2}} - \frac{1 - u^{2}}{1 + u^{2}})}$ $= \int_{0}^{\infty} \frac{2 d u}{3 + 3 u^{2} - 4 u - 1 + u^{2}} = \int_{0}^{\infty} \frac{2 d u}{4 u^{2} - 4 u + 2}$ $= \int_{0}^{\infty} \frac{d u}{2 u^{2} - 2 u + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{d u}{u^{2} - u + \frac{1}{2}} = \frac{1}{2} \int_{0}^{\infty} \frac{d u}{{(u - \frac{1}{2})}^{2} + \frac{3}{4} + \frac{1}{2}}$ $= \frac{1}{2} \int_{0}^{\infty} \frac{d u}{{(u - \frac{1}{2})}^{2} + \frac{5}{4}} =_{u - \frac{1}{2} = \frac{\sqrt{5}}{2} α} \frac{1}{2} \times \frac{4}{5} \int_{- \frac{1}{\sqrt{5}}}^{+ \infty} \frac{1}{1 + u^{2}} \frac{\sqrt{5}}{2} d α$ $= \frac{1}{\sqrt{5}} {[a r c t a n α]}_{- \frac{1}{\sqrt{5}}}^{+ \infty} = \frac{1}{\sqrt{5}} {\frac{π}{2} + a r c t a n (\frac{1}{\sqrt{5}})} \Rightarrow$ $I = \frac{π}{4} + \frac{π}{2 \sqrt{5}} + \frac{1}{\sqrt{5}} a r c t a n (\frac{1}{\sqrt{5}}) .$
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