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Question Number 101079 by I want to learn more last updated on 30/Jun/20

Commented by I want to learn more last updated on 30/Jun/20

$Expecially question (2) please$
Commented by Dwaipayan Shikari last updated on 01/Jul/20

$\int_{- π}^{π} \frac{d x}{2 - (\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}})} . \frac{\frac{1}{2} {s e c}^{2} \frac{x}{2}}{\frac{1}{2} {s e c}^{2} \frac{x}{2}} = 2 \int_{- π}^{π} \frac{d t}{(1 + t^{2}) (\frac{2 + 2 t^{2} - 1 + t^{2} - 2 t}{1 + t^{2}})}$ $S u p p o s e t a n x = t a n d \frac{d t}{d x} = \frac{1}{2} {s e c}^{2} x a n d {s e c}^{2} \frac{x}{2} = 1 + t^{2}$ $= 2 \int_{- π}^{π} \frac{d t}{3 t^{2} - 2 t + 1} = 2 \int_{- π}^{π} \frac{d t}{{(3 t - \frac{1}{3})}^{2} + {(\frac{2 \sqrt{2}}{3})}^{2}} = \frac{2}{3} \int_{- π}^{π} \frac{d u}{u^{2} + {(\frac{2 \sqrt{2}}{3})}^{2}}$ $= \frac{2}{3} \frac{3}{2 \sqrt{2}} \tan^{- 1} (\frac{3 u}{2 \sqrt{2}}) = \sqrt{2} {t a n}^{- 1} (\frac{9 t - 1}{2 \sqrt{2}}) = {[\sqrt{2} {t a n}^{- 1} (\frac{9 t a n \frac{x}{2} - 1}{2 \sqrt{2}})]}_{- π}^{π}$ $s u p p o s e (3 t - \frac{1}{3}) = u$ $= \sqrt{2} (\frac{π}{2} + \frac{π}{2}) = \sqrt{2} π ★$
	Answered by abdomathmax last updated on 01/Jul/20
	$classic method A =∫_(−π) ^π (dx/(2−(cosx +sinx))) changement tan((x/2))=t give A =∫_(−∞) ^(+∞) ((2dt)/((1+t^2 ){2−(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 )))})) =∫_(−∞) ^(+∞ ) ((2dt)/(2+2t^2 −1+t^2 −2t)) =∫_(−∞) ^(+∞) ((2dt)/(3t^2 −2t+1)) =(2/3) ∫_(−∞) ^(+∞) (dt/(t^2 −(2/3)t +(1/3))) =(2/3) ∫_(−∞) ^(+∞ ) (dt/(t^2 −(2/3)t +(1/9)+(1/3)−(1/9))) =(2/3) ∫_(−∞) ^(+∞) (dt/((t−(1/3))^2 +(2/9))) =_(t−(1/3)=((√2)/3)u) (2/3)×(9/2) ∫_(−∞) ^(+∞) (1/(u^2 +1))×((√2)/3) du =(√2)∫_(−∞) ^(+∞) (du/(1+u^2 )) =(√2)[ arctanu]_(−∞) ^(+∞) =(√2){(π/2)−(−(π/2))} =π(√2) ★A =π(√2)★$
	$classic method A = \int_{- π}^{π} \frac{dx}{2 - (cosx + sinx)}$ $changement \tan (\frac{x}{2}) = t give$ $A = \int_{- \infty}^{+ \infty} \frac{2 dt}{(1 + t^{2}) {2 - (\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}})}}$ $= \int_{- \infty}^{+ \infty} \frac{2 dt}{2 + {2 t}^{2} - 1 + t^{2} - 2 t} = \int_{- \infty}^{+ \infty} \frac{2 dt}{{3 t}^{2} - 2 t + 1}$ $= \frac{2}{3} \int_{- \infty}^{+ \infty} \frac{dt}{t^{2} - \frac{2}{3} t + \frac{1}{3}}$ $= \frac{2}{3} \int_{- \infty}^{+ \infty} \frac{dt}{t^{2} - \frac{2}{3} t + \frac{1}{9} + \frac{1}{3} - \frac{1}{9}}$ $= \frac{2}{3} \int_{- \infty}^{+ \infty} \frac{dt}{{(t - \frac{1}{3})}^{2} + \frac{2}{9}}$ $=_{t - \frac{1}{3} = \frac{\sqrt{2}}{3} u} \frac{2}{3} \times \frac{9}{2} \int_{- \infty}^{+ \infty} \frac{1}{u^{2} + 1} \times \frac{\sqrt{2}}{3} du$ $= \sqrt{2} \int_{- \infty}^{+ \infty} \frac{du}{1 + u^{2}} = \sqrt{2} {[arctanu]}_{- \infty}^{+ \infty}$ $= \sqrt{2} {\frac{π}{2} - (- \frac{π}{2})} = π \sqrt{2}$ $★ A = π \sqrt{2} ★$
	Commented by I want to learn more last updated on 02/Jul/20

	$Thanks sir$
	Answered by mr W last updated on 30/Jun/20
	$∫_(−π) ^π (dx/(2−(cos x+sin x))) =(1/(√2))∫_(−π) ^π (dx/((√2)−cos (x−(π/4)))) =(1/(√2))∫_(−((5π)/4)) ^((3π)/4) (dt/((√2)−cos t)) =(√2)[tan^(−1) {((√2)+1)tan (t/2)}]_(−((5π)/4)) ^(−π) +(√2)[tan^(−1) {((√2)+1)tan (t/2)}]_(−π) ^((3π)/4) =(√2)[(π/2)−tan^(−1) ((√2)+1)^2 +tan^(−1) ((√2)+1)^2 +(π/2)] =(√2)π$
	$\int_{- π}^{π} \frac{d x}{2 - (\cos x + \sin x)}$ $= \frac{1}{\sqrt{2}} \int_{- π}^{π} \frac{d x}{\sqrt{2} - \cos (x - \frac{π}{4})}$ $= \frac{1}{\sqrt{2}} \int_{- \frac{5 π}{4}}^{\frac{3 π}{4}} \frac{d t}{\sqrt{2} - \cos t}$ $= \sqrt{2} {[\tan^{- 1} {(\sqrt{2} + 1) \tan \frac{t}{2}}]}_{- \frac{5 π}{4}}^{- π}$ $+ \sqrt{2} {[\tan^{- 1} {(\sqrt{2} + 1) \tan \frac{t}{2}}]}_{- π}^{\frac{3 π}{4}}$ $= \sqrt{2} [\frac{π}{2} - \tan^{- 1} {(\sqrt{2} + 1)}^{2} + \tan^{- 1} {(\sqrt{2} + 1)}^{2} + \frac{π}{2}]$ $= \sqrt{2} π$
	Commented by I want to learn more last updated on 30/Jun/20

	$Thanks sir . I appreciate$
	Commented by mr W last updated on 30/Jun/20

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