∫_0 ^( (π/4)) ((sin(ς)dς)/(cos(ς)+sin(ς)))dς where ς : zeta


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Question Number 124963 by 676597498 last updated on 07/Dec/20

$\int_{0}^{\frac{π}{4}} \frac{\sin (ς) d ς}{\cos (ς) + \sin (ς)} d ς$ $where ς : zeta$
Commented by MJS_new last updated on 07/Dec/20

$syntax error .$
	Answered by Dwaipayan Shikari last updated on 07/Dec/20

	$\int_{0}^{\frac{π}{4}} \frac{1}{t a n (ζ) + 1} d ζ = \int_{0}^{1} \frac{1}{t + 1} . \frac{1}{t^{2} + 1} d t t a n ζ = t \Rightarrow {s e c}^{2} ζ = \frac{d t}{d ζ}$ $= \frac{1}{2} \int_{0}^{1} \frac{1}{t + 1} - \frac{t - 1}{t^{2} + 1} d t$ $= \frac{1}{2} l o g (2) - \frac{1}{4} \int_{0}^{1} \frac{2 t}{t^{2} + 1} + \frac{1}{2} \int_{0}^{1} \frac{1}{1 + t^{2}} d t$ $= - \frac{1}{4} l o g (2) + \frac{π}{8}$
	Commented by 676597498 last updated on 07/Dec/20

	$thanks$ $but there are 2 (d ς)$ $look again$ $it is \int \frac{\sin (ς) d ς}{\cos (ς) + \sin (ς)} d ς$
	Commented by Dwaipayan Shikari last updated on 07/Dec/20

	$S o r r y b u t i t {d o e s n}^{'} t m a k e a n y s e n s e t o m e$
	Commented by 676597498 last updated on 07/Dec/20

	$it makes sense$
	Commented by mr W last updated on 07/Dec/20

	$it makes N o sense!$
	Commented by MJS_new last updated on 07/Dec/20

	$if it makes sense, what sense does it make ?$
	Answered by mathmax by abdo last updated on 07/Dec/20
	$A =∫_0 ^(π/4) ((sinx)/(cosx +sinx))dx we do the changement tan((x/2))=t ⇒ A =∫_0 ^((√2)−1) (((2t)/(1+t^2 ))/(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 ))))((2dt)/(1+t^2 )) =∫_0 ^((√2)−1) ((4t)/((t^2 +1)(−t^2 +2t+1)))dt =−4 ∫_0 ^((√2)−1) ((tdt)/((t^2 +1)(t^2 −2t−1))) let decompose F(t)=(t/((t^2 +1)(t^2 −2t−1))) t^2 −2t−1→Δ^′ =1+1=2 ⇒t_1 =1+(√2) and t_2 =1−(√2) ⇒ F(t)=(t/((t−t_1 )(t−t_2 )(t^2 +1))) =(a/(t−t_1 )) +(b/(t−t_2 )) +((mt +n)/(t^2 +1)) a=(t_1 /(2(√2)(t_1 ^2 +1))) =((1+(√2))/(2(√2)(3+2(√2)+1)))=((1+(√2))/(2(√2)(4+2(√2))))=((1+(√2))/(8(√2)+8))=(1/8) b =(t_2 /(−2(√2)(t_2 ^2 +1))) =((1−(√2))/(−2(√2)(3−2(√2)+1)))=((1−(√2))/(−2(√2)(4−2(√2)))) =((1−(√2))/(−8(√2)+8)) =(1/8) lim_(t→+∞) tF(t)=0=a+b+m ⇒m=−(1/4) F(o)=0 =−(a/t_1 )−(b/t_2 ) +n ⇒n=(a/t_1 )+(b/t_2 ) =(1/(8(1+(√2))))+(1/(8(1−(√2)))) =(1/8)((2/((−1))))=−(1/4) ⇒ F(t)=(1/(8(t−(1+(√2))))+(1/(8(t−(1−(√2)))))−(1/4)×((t+1)/(t^2 +1)) ⇒ ∫_0 ^((√2)−1) F(t)dt =(1/8)∫_0 ^((√2)−1) (dt/(t−1−(√2))) +(1/8)∫_0 ^((√2)−1) (dt/(t−1+(√2))) −(1/8)∫_0 ^((√2)−1) ((2t)/(t^2 +1))−(1/4)∫_0 ^((√2)−1) (dt/(t^2 +1)) =(1/8)[ln∣(t−1)^2 −2∣]_0 ^((√2)−1) −(1/8)[ln(t^2 +1)]_0 ^((√2)−1) −(1/4)[arctant]_0 ^((√2)−1) =(1/8){ln∣((√2)−2)^2 −2∣−(1/8)ln(((√2)−1)^2 +1)−(1/4)×(π/8) =(1/8)ln∣2−4(√2)+4−2∣−(1/8)ln(3−2(√2)+1)−(π/(32)) =(1/8)ln(4(√2)−4)−(1/8)ln(4−2(√2))−(π/(32)) ⇒ I=−(1/2)ln(4(√2)−4)+(1/2)ln(4−2(√2))+(π/8)$
	$A = \int_{0}^{\frac{π}{4}} \frac{sinx}{cosx + sinx} dx we do the changement \tan (\frac{x}{2}) = t \Rightarrow$ $A = \int_{0}^{\sqrt{2} - 1} \frac{\frac{2 t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \frac{2 dt}{1 + t^{2}} = \int_{0}^{\sqrt{2} - 1} \frac{4 t}{(t^{2} + 1) (- t^{2} + 2 t + 1)} dt$ $= - 4 \int_{0}^{\sqrt{2} - 1} \frac{tdt}{(t^{2} + 1) (t^{2} - 2 t - 1)} let decompose$ $F (t) = \frac{t}{(t^{2} + 1) (t^{2} - 2 t - 1)}$ $t^{2} - 2 t - 1 \to Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} and t_{2} = 1 - \sqrt{2} \Rightarrow$ $F (t) = \frac{t}{(t - t_{1}) (t - t_{2}) (t^{2} + 1)} = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{mt + n}{t^{2} + 1}$ $a = \frac{t_{1}}{2 \sqrt{2} (t_{1}^{2} + 1)} = \frac{1 + \sqrt{2}}{2 \sqrt{2} (3 + 2 \sqrt{2} + 1)} = \frac{1 + \sqrt{2}}{2 \sqrt{2} (4 + 2 \sqrt{2})} = \frac{1 + \sqrt{2}}{8 \sqrt{2} + 8} = \frac{1}{8}$ $b = \frac{t_{2}}{- 2 \sqrt{2} (t_{2}^{2} + 1)} = \frac{1 - \sqrt{2}}{- 2 \sqrt{2} (3 - 2 \sqrt{2} + 1)} = \frac{1 - \sqrt{2}}{- 2 \sqrt{2} (4 - 2 \sqrt{2})}$ $= \frac{1 - \sqrt{2}}{- 8 \sqrt{2} + 8} = \frac{1}{8}$ $\lim_{t \to + \infty} tF (t) = 0 = a + b + m \Rightarrow m = - \frac{1}{4}$ $F (o) = 0 = - \frac{a}{t_{1}} - \frac{b}{t_{2}} + n \Rightarrow n = \frac{a}{t_{1}} + \frac{b}{t_{2}} = \frac{1}{8 (1 + \sqrt{2})} + \frac{1}{8 (1 - \sqrt{2})}$ $= \frac{1}{8} (\frac{2}{(- 1)}) = - \frac{1}{4} \Rightarrow$ $F (t) = \frac{1}{8 (t - (1 + \sqrt{2})} + \frac{1}{8 (t - (1 - \sqrt{2}))} - \frac{1}{4} \times \frac{t + 1}{t^{2} + 1} \Rightarrow$ $\int_{0}^{\sqrt{2} - 1} F (t) dt = \frac{1}{8} \int_{0}^{\sqrt{2} - 1} \frac{dt}{t - 1 - \sqrt{2}} + \frac{1}{8} \int_{0}^{\sqrt{2} - 1} \frac{dt}{t - 1 + \sqrt{2}}$ $- \frac{1}{8} \int_{0}^{\sqrt{2} - 1} \frac{2 t}{t^{2} + 1} - \frac{1}{4} \int_{0}^{\sqrt{2} - 1} \frac{dt}{t^{2} + 1}$ $= \frac{1}{8} {[\ln ∣ {(t - 1)}^{2} - 2 ∣]}_{0}^{\sqrt{2} - 1} - \frac{1}{8} {[\ln (t^{2} + 1)]}_{0}^{\sqrt{2} - 1} - \frac{1}{4} {[arctant]}_{0}^{\sqrt{2} - 1}$ $= \frac{1}{8} {\ln ∣ {(\sqrt{2} - 2)}^{2} - 2 ∣ - \frac{1}{8} \ln ({(\sqrt{2} - 1)}^{2} + 1) - \frac{1}{4} \times \frac{π}{8}$ $= \frac{1}{8} \ln ∣ 2 - 4 \sqrt{2} + 4 - 2 ∣ - \frac{1}{8} \ln (3 - 2 \sqrt{2} + 1) - \frac{π}{32}$ $= \frac{1}{8} \ln (4 \sqrt{2} - 4) - \frac{1}{8} \ln (4 - 2 \sqrt{2}) - \frac{π}{32} \Rightarrow$ $I = - \frac{1}{2} \ln (4 \sqrt{2} - 4) + \frac{1}{2} \ln (4 - 2 \sqrt{2}) + \frac{π}{8}$
	Commented by talminator2856791 last updated on 07/Dec/20

	$abdo to the rescue!!!!!$
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