find ∫_0 ^(π/4) (dx/(2cosx +cos(2x)))


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Question Number 37784 by prof Abdo imad last updated on 17/Jun/18

$f i n d \int_{0}^{\frac{π}{4}} \frac{d x}{2 c o s x + c o s (2 x)}$
Commented by math khazana by abdo last updated on 18/Jun/18
$I = ∫_0 ^(π/4) (dx/(2cosx +2 cos^2 x−1)) =∫_0 ^(π/4) (dx/(2cos^2 x +2cosx −1)) let decompose F(x)= (1/(2x^2 +2x −1)) Δ^′ =1+2=3 ⇒x_1 =((−1+(√3))/2) and x_2 =((−1−(√3))/2) ⇒ F(x)= (1/(2(x−x_1 )(x−x_2 ))) =(a/(x−x_1 )) +(b/(x−x_2 )) a= (1/(2(x_1 −x_2 ))) =(1/(2.(√3))) b= (1/(2(x_2 −x_1 ))) = (1/(2(−(√3)))) =−(1/(2(√3))) ⇒ F(x)= (1/(2(√3))){ (1/(x −((−1+(√3))/2))) −(1/(x−((−1−(√3))/2)))} = (1/(2(√3))){ (2/(2x+1−(√3))) − (2/(2x +1+(√3)))} ⇒ I = (1/(√3)) ∫_0 ^(π/4) (dx/(2cosx +1−(√3))) −(1/(√3))∫_0 ^(π/4) (dx/(2cosx +1+(√3))) =H −K changement tan((x/2))=t give H = (1/(√3)) ∫_0 ^((√2)−1) (1/(2((1−t^2 )/(1+t^2 )) +1−(√3))) ((2dt)/(1+t^2 )) = (2/(√3)) ∫_0 ^((√2)−1) (dt/(2−2t^2 +1−(√3) +(1−(√3))t^2 )) =(2/(√3)) ∫_0 ^((√2)−1) (dt/(3−(√3) −(1+(√3))t^2 )) =(2/((√3)(3−(√3)))) ∫_0 ^((√2)−1) (dt/(1−((√((1+(√3))/(3−(√3))))t)^2 )) =_((√((1+(√3))/(3−(√3))))t=u) (2/((√3)(3−(√3)))) ∫_0 ^(((√2)−1)(√((1+(√3))/(3−(√3))))) (1/(1−u^2 )) (√((3−(√3))/(1+(√3))))du = (2/((√3)(√((3−(√(3)(1+(√(3))))))))) ∫_0 ^(((√(2−1)))(√((1+(√3))/(3−(√3))))) (du/(1−u^2 )) = (1/((√3)(√((3−(√3))(1+(√3))))))[ln∣ ((1+u)/(1−u))∣]_0 ^(((√2)−1)(√((1+(√3))/(3−(√3))))) = ((√3)/(3(√((3−(√3))(1+(√3))))))ln( ((1+λ)/(1−λ))) with λ= ((√2)−1)(√((1+(√3))/(3−(√3)))) we follow the same?method to calculate K....$
$I = \int_{0}^{\frac{π}{4}} \frac{d x}{2 c o s x + 2 {c o s}^{2} x - 1}$ $= \int_{0}^{\frac{π}{4}} \frac{d x}{2 {c o s}^{2} x + 2 c o s x - 1} l e t d e c o m p o s e$ $F (x) = \frac{1}{2 x^{2} + 2 x - 1}$ $Δ^{^{'}} = 1 + 2 = 3 \Rightarrow x_{1} = \frac{- 1 + \sqrt{3}}{2} a n d x_{2} = \frac{- 1 - \sqrt{3}}{2} \Rightarrow$ $F (x) = \frac{1}{2 (x - x_{1}) (x - x_{2})} = \frac{a}{x - x_{1}} + \frac{b}{x - x_{2}}$ $a = \frac{1}{2 (x_{1} - x_{2})} = \frac{1}{2 . \sqrt{3}}$ $b = \frac{1}{2 (x_{2} - x_{1})} = \frac{1}{2 (- \sqrt{3})} = - \frac{1}{2 \sqrt{3}} \Rightarrow$ $F (x) = \frac{1}{2 \sqrt{3}} {\frac{1}{x - \frac{- 1 + \sqrt{3}}{2}} - \frac{1}{x - \frac{- 1 - \sqrt{3}}{2}}}$ $= \frac{1}{2 \sqrt{3}} {\frac{2}{2 x + 1 - \sqrt{3}} - \frac{2}{2 x + 1 + \sqrt{3}}} \Rightarrow$ $I = \frac{1}{\sqrt{3}} \int_{0}^{\frac{π}{4}} \frac{d x}{2 c o s x + 1 - \sqrt{3}} - \frac{1}{\sqrt{3}} \int_{0}^{\frac{π}{4}} \frac{d x}{2 c o s x + 1 + \sqrt{3}}$ $= H - K$ $c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $H = \frac{1}{\sqrt{3}} \int_{0}^{\sqrt{2} - 1} \frac{1}{2 \frac{1 - t^{2}}{1 + t^{2}} + 1 - \sqrt{3}} \frac{2 d t}{1 + t^{2}}$ $= \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{2} - 1} \frac{d t}{2 - 2 t^{2} + 1 - \sqrt{3} + (1 - \sqrt{3}) t^{2}}$ $= \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{2} - 1} \frac{d t}{3 - \sqrt{3} - (1 + \sqrt{3}) t^{2}}$ $= \frac{2}{\sqrt{3} (3 - \sqrt{3})} \int_{0}^{\sqrt{2} - 1} \frac{d t}{1 - {(\sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}} t)}^{2}}$ $=_{\sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}} t = u} \frac{2}{\sqrt{3} (3 - \sqrt{3})} \int_{0}^{(\sqrt{2} - 1) \sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}}} \frac{1}{1 - u^{2}} \sqrt{\frac{3 - \sqrt{3}}{1 + \sqrt{3}}} d u$ $= \frac{2}{\sqrt{3} \sqrt{(3 - \sqrt{3) (1 + \sqrt{3)}}}} \int_{0}^{(\sqrt{2 - 1}) \sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}}} \frac{d u}{1 - u^{2}}$ $= \frac{1}{\sqrt{3} \sqrt{(3 - \sqrt{3}) (1 + \sqrt{3})}} {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{(\sqrt{2} - 1) \sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}}}$ $= \frac{\sqrt{3}}{3 \sqrt{(3 - \sqrt{3}) (1 + \sqrt{3})}} l n (\frac{1 + λ}{1 - λ}) w i t h$ $λ = (\sqrt{2} - 1) \sqrt{\frac{1 + \sqrt{3}}{3 - \sqrt{3}}}$ $w e f o l l o w t h e s a m e ? m e t h o d t o c a l c u l a t e K . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18
	$∫_0 ^(Π/4) (dx/(2cosx+2cos^2 x−1)) =(1/2)∫_0 ^(Π/4) (dx/(cos^2 x+cosx−(1/2))) a^2 +a−(1/2) a^2 +2.a.(1/2)+(1/4)−(1/4)−(1/2) (a+(1/2))^2 −(3/4) (a+(1/2)+(((√3) )/2))(a+(1/2)−(((√3) )/2)) (a+((1+(√3) )/2))(a+((1−(√3) )/2)) (a+((1+(√3) )/2))−(a+((1−(√3) )/2)) (√3) =(1/(2(√3)))∫_0 ^(Π/4) (((√3) )/((cosx+((1+(√3) )/2))(cosx+((1−(√3) )/2))))dx =(1/(2(√3) ))∫_0 ^(Π/4) (dx/(cosx+((1−(√3) )/2)))−(1/(2(√3)))∫_0 ^(Π/4) (dx/(cosx+((1−(√3) )/2))) =(1/(2(√3) )){∫_0 ^(Π/4) (dx/(a+cosx))−∫_0 ^(Π/4) (dx/(b+cosx))} now it can be solved but lenthy...method...∫ ∫(dx/(a+cosx)) ∫(dx/(a+((1−tan^2 (x/2))/(1+tan^2 (x/2))))) ∫((sec^2 (x/2))/(a+atan^2 (x/2)+1−tan^2 (x/2)))dx ∫((sec^2 (x/2))/((a+1)+tan^2 (x/2)(a−1)))dx (1/(a−1))∫((sec^2 (x/2))/(((a+1)/(a−1))+tan^2 (x/2)))dx t=tan(x/2) dt=sec^2 (x/2)×(1/2)dx (2/(a−1))∫(dt/({(√(((a+1)/(a−1)) )) }^2 +t^2 )) (2/(a−1))×(((√(a−1)) )/(√(a+1))) ×tan^(−1) {(t/(((√(a−1)) )/(√(a+1))))}...like this way$
	$\int_{0}^{\frac{Π}{4}} \frac{d x}{2 c o s x + 2 {c o s}^{2} x - 1}$ $= \frac{1}{2} \int_{0}^{\frac{Π}{4}} \frac{d x}{{c o s}^{2} x + c o s x - \frac{1}{2}}$ $a^{2} + a - \frac{1}{2}$ $a^{2} + 2 . a . \frac{1}{2} + \frac{1}{4} - \frac{1}{4} - \frac{1}{2}$ ${(a + \frac{1}{2})}^{2} - \frac{3}{4}$ $(a + \frac{1}{2} + \frac{\sqrt{3}}{2}) (a + \frac{1}{2} - \frac{\sqrt{3}}{2})$ $(a + \frac{1 + \sqrt{3}}{2}) (a + \frac{1 - \sqrt{3}}{2})$ $(a + \frac{1 + \sqrt{3}}{2}) - (a + \frac{1 - \sqrt{3}}{2})$ $\sqrt{3}$ $= \frac{1}{2 \sqrt{3}} \int_{0}^{\frac{Π}{4}} \frac{\sqrt{3}}{(c o s x + \frac{1 + \sqrt{3}}{2}) (c o s x + \frac{1 - \sqrt{3}}{2})} d x$ $= \frac{1}{2 \sqrt{3}} \int_{0}^{\frac{Π}{4}} \frac{d x}{c o s x + \frac{1 - \sqrt{3}}{2}} - \frac{1}{2 \sqrt{3}} \int_{0}^{\frac{Π}{4}} \frac{d x}{c o s x + \frac{1 - \sqrt{3}}{2}}$ $= \frac{1}{2 \sqrt{3}} {\int_{0}^{\frac{Π}{4}} \frac{d x}{a + c o s x} - \int_{0}^{\frac{Π}{4}} \frac{d x}{b + c o s x}}$ $n o w i t c a n b e s o l v e d b u t l e n t h y . . . m e t h o d . . . \int$ $\int \frac{d x}{a + c o s x}$ $\int \frac{d x}{a + \frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}}$ $\int \frac{{s e c}^{2} \frac{x}{2}}{a + {a t a n}^{2} \frac{x}{2} + 1 - {t a n}^{2} \frac{x}{2}} d x$ $\int \frac{{s e c}^{2} \frac{x}{2}}{(a + 1) + {t a n}^{2} \frac{x}{2} (a - 1)} d x$ $\frac{1}{a - 1} \int \frac{{s e c}^{2} \frac{x}{2}}{\frac{a + 1}{a - 1} + {t a n}^{2} \frac{x}{2}} d x$ $t = t a n \frac{x}{2} d t = {s e c}^{2} \frac{x}{2} \times \frac{1}{2} d x$ $\frac{2}{a - 1} \int \frac{d t}{{\sqrt{\frac{a + 1}{a - 1}}}^{2} + t^{2}}$ $\frac{2}{a - 1} \times \frac{\sqrt{a - 1}}{\sqrt{a + 1}} \times {t a n}^{- 1} {\frac{t}{\frac{\sqrt{a - 1}}{\sqrt{a + 1}}}} . . . l i k e t h i s w a y$
	Answered by MJS last updated on 17/Jun/18

	$Weierstrass$ $\int \frac{d x}{2 \cos x + \cos 2 x} =$ $[t = \tan \frac{x}{2} \to d x = \frac{2 d t}{1 + t^{2}}]$ $= - 2 \int \frac{t^{2} + 1}{t^{4} + 6 t^{2} - 3} d t =$ $= - 2 \int \frac{t^{2} + 1}{(t - \sqrt{- 3 + 2 \sqrt{3}}) (t + \sqrt{- 3 + 2 \sqrt{3}}) (t^{2} + 3 + 2 \sqrt{3})} d t =$ $= - 2 (\int \frac{A}{t - \sqrt{- 3 + 2 \sqrt{3}}} d t + \int \frac{B}{t + \sqrt{- 3 + 2 \sqrt{3}}} d t + \int \frac{C}{t^{2} + 3 + 2 \sqrt{3}} d t) =$ $= - \frac{\sqrt{2 \sqrt{3}}}{6} \int \frac{d t}{t - \sqrt{- 3 + 2 \sqrt{3}}} + \frac{\sqrt{2 \sqrt{3}}}{6} \int \frac{d t}{t + \sqrt{- 3 + 2 \sqrt{3}}} - \frac{3 + \sqrt{3}}{3} \int \frac{d t}{t^{2} + 3 + 2 \sqrt{3}}) =$ $[\int \frac{d x}{x \pm a} = \ln ∣ x \pm a ∣; \int \frac{d x}{x^{2} + a} = \frac{1}{\sqrt{a}} \arctan \frac{x}{\sqrt{a}}]$ $= \frac{\sqrt{2 \sqrt{3}}}{6} (\ln ∣ t + \sqrt{- 3 + 2 \sqrt{3}} ∣ - \ln ∣ t - \sqrt{- 3 + 2 \sqrt{3}} ∣) - (1 + \frac{\sqrt{3}}{3}) \frac{1}{\sqrt{3 + 2 \sqrt{3}}} \arctan \frac{t}{\sqrt{3 + 2 \sqrt{3}}} =$ $= \frac{\sqrt{2 \sqrt{3}}}{6} \ln ∣ \frac{t + \sqrt{- 3 + 2 \sqrt{3}}}{t - \sqrt{- 3 + 2 \sqrt{3}}} ∣ - \frac{\sqrt{2 \sqrt{3}}}{3} \arctan \frac{\sqrt{2 \sqrt{3}} (3 - \sqrt{3}) t}{6} =$ $= \frac{\sqrt{2 \sqrt{3}}}{6} (\ln ∣ \frac{t + \sqrt{- 3 + 2 \sqrt{3}}}{t - \sqrt{- 3 + 2 \sqrt{3}}} ∣ - 2 \arctan \frac{\sqrt{2 \sqrt{3}} (3 - \sqrt{3}) t}{6}) =$ $= \frac{\sqrt{2 \sqrt{3}}}{6} (\ln ∣ \frac{\tan \frac{x}{2} + \sqrt{- 3 + 2 \sqrt{3}}}{\tan \frac{x}{2} - \sqrt{- 3 + 2 \sqrt{3}}} ∣ - 2 \arctan \frac{\sqrt{2 \sqrt{3}} (3 - \sqrt{3}) \tan \frac{x}{2}}{6}) + C$
	Commented by math khazana by abdo last updated on 18/Jun/18

	$g o o d h a r d w o r k$
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