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Question Number 124963 by 676597498 last updated on 07/Dec/20
∫_0 ^( (π/4)) ((sin(ς)dς)/(cos(ς)+sin(ς)))dς where ς : zeta
$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sin}\left(\varsigma\right)\mathrm{d}\varsigma}{\mathrm{cos}\left(\varsigma\right)+\mathrm{sin}\left(\varsigma\right)}\mathrm{d}\varsigma \\ $$$$\mathrm{where}\:\varsigma\::\:\mathrm{zeta}\: \\ $$
Commented by MJS_new last updated on 07/Dec/20
syntax error.
$$\mathrm{syntax}\:\mathrm{error}. \\ $$
Answered by Dwaipayan Shikari last updated on 07/Dec/20
∫_0 ^(π/4) (1/(tan(ζ)+1))dζ =∫_0 ^1 (1/(t+1)).(1/(t^2 +1))dt tanζ=t⇒sec^2 ζ=(dt/dζ) =(1/2)∫_0 ^1 (1/(t+1))−((t−1)/(t^2 +1))dt =(1/2)log(2)−(1/4)∫_0 ^1 ((2t)/(t^2 +1))+(1/2)∫_0 ^1 (1/(1+t^2 ))dt =−(1/4)log(2)+(π/8)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}}{{tan}\left(\zeta\right)+\mathrm{1}}{d}\zeta\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}+\mathrm{1}}.\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}\zeta={t}\Rightarrow{sec}^{\mathrm{2}} \zeta=\frac{{dt}}{{d}\zeta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}{log}\left(\mathrm{2}\right)+\frac{\pi}{\mathrm{8}} \\ $$
Commented by 676597498 last updated on 07/Dec/20
thanks but there are 2 (dς) look again it is ∫((sin(ς)dς)/(cos(ς)+sin(ς)))dς
$$\mathrm{thanks} \\ $$$$\mathrm{but}\:\mathrm{there}\:\mathrm{are}\:\mathrm{2}\:\left(\mathrm{d}\varsigma\right) \\ $$$$\mathrm{look}\:\mathrm{again} \\ $$$$\mathrm{it}\:\mathrm{is}\:\int\frac{\mathrm{sin}\left(\varsigma\right)\mathrm{d}\varsigma}{\mathrm{cos}\left(\varsigma\right)+\mathrm{sin}\left(\varsigma\right)}\mathrm{d}\varsigma \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 07/Dec/20
Sorry but it doesn′t make any sense to me
$${Sorry}\:{but}\:{it}\:{doesn}'{t}\:{make}\:{any}\:{sense}\:{to}\:{me} \\ $$
Commented by 676597498 last updated on 07/Dec/20
it makes sense
$$\mathrm{it}\:\mathrm{makes}\:\mathrm{sense} \\ $$
Commented by mr W last updated on 07/Dec/20
it makes No sense!
$$\mathrm{it}\:\mathrm{makes}\:\boldsymbol{{No}}\:\mathrm{sense}! \\ $$
Commented by MJS_new last updated on 07/Dec/20
if it makes sense, what sense does it make?
$$\mathrm{if}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{sense},\:\mathrm{what}\:\mathrm{sense}\:\mathrm{does}\:\mathrm{it}\:\mathrm{make}? \\ $$
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)
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{sinx}}{\mathrm{cosx}\:+\mathrm{sinx}}\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\frac{\mathrm{2dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{\mathrm{4t}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(−\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}+\mathrm{1}\right)}\mathrm{dt} \\ $$$$=−\mathrm{4}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{tdt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2t}−\mathrm{1}\right)}\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2t}−\mathrm{1}\right)} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{2t}−\mathrm{1}\rightarrow\Delta^{'} \:=\mathrm{1}+\mathrm{1}=\mathrm{2}\:\Rightarrow\mathrm{t}_{\mathrm{1}} =\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{t}_{\mathrm{2}} =\mathrm{1}−\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{\left(\mathrm{t}−\mathrm{t}_{\mathrm{1}} \right)\left(\mathrm{t}−\mathrm{t}_{\mathrm{2}} \right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\mathrm{a}}{\mathrm{t}−\mathrm{t}_{\mathrm{1}} }\:+\frac{\mathrm{b}}{\mathrm{t}−\mathrm{t}_{\mathrm{2}} }\:+\frac{\mathrm{mt}\:+\mathrm{n}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{a}=\frac{\mathrm{t}_{\mathrm{1}} }{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{t}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}\right)}=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\right)}=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{b}\:=\frac{\mathrm{t}_{\mathrm{2}} }{−\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{t}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{−\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}\right)}=\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{−\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{−\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{8}}\:=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{lim}_{\mathrm{t}\rightarrow+\infty} \:\mathrm{tF}\left(\mathrm{t}\right)=\mathrm{0}=\mathrm{a}+\mathrm{b}+\mathrm{m}\:\Rightarrow\mathrm{m}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{F}\left(\mathrm{o}\right)=\mathrm{0}\:=−\frac{\mathrm{a}}{\mathrm{t}_{\mathrm{1}} }−\frac{\mathrm{b}}{\mathrm{t}_{\mathrm{2}} }\:+\mathrm{n}\:\Rightarrow\mathrm{n}=\frac{\mathrm{a}}{\mathrm{t}_{\mathrm{1}} }+\frac{\mathrm{b}}{\mathrm{t}_{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{2}}{\left(−\mathrm{1}\right)}\right)=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{t}−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right.}+\frac{\mathrm{1}}{\mathrm{8}\left(\mathrm{t}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\right)}−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{t}+\mathrm{1}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}−\mathrm{1}−\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}−\mathrm{1}+\sqrt{\mathrm{2}}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{ln}\mid\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\mid\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{arctant}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left\{\mathrm{ln}\mid\left(\sqrt{\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\mid−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\left(\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\pi}{\mathrm{8}}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\mid\mathrm{2}−\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{4}−\mathrm{2}\mid−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}\right)−\frac{\pi}{\mathrm{32}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{4}\right)−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)−\frac{\pi}{\mathrm{32}}\:\Rightarrow \\ $$$$\mathrm{I}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{4}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)+\frac{\pi}{\mathrm{8}} \\ $$
Commented by talminator2856791 last updated on 07/Dec/20
abdo to the rescue!!!!!
$$\:\mathrm{abdo}\:\mathrm{to}\:\mathrm{the}\:\mathrm{rescue}!!!!! \\ $$

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