Question-80846

Question Number 80846 by TawaTawa last updated on 07/Feb/20

Commented by mathmax by abdo last updated on 07/Feb/20

we havw ∫_(−π) ^π costdt =2∫_0 ^π cost dt =2[sint]_0 ^π =0 ∫_0 ^(π/2) ((cosθ)/(2−sin(2θ)))dθ =(1/2) ∫_0 ^(π/2) ((cosθ)/(1−cosθ sinθ))dθ=_(tan((θ/2))=t) =(1/2) ∫_0 ^1 (((1−t^2 )/(1+t^2 ))/(1−((1−t^2 )/(1+t^2 ))×((2t)/(1+t^2 ))))×((2dt)/(1+t^2 )) =∫_0 ^1 ((1−t^2 )/((1+t^2 )^2 {1−((2t(1−t^2 ))/((1+t^2 )^2 ))}))dt =∫_0 ^1 ((1−t^2 )/((1+t^2 )^2 −2t+2t^3 ))dt =∫_0 ^1 ((1−t^2 )/(t^4 +2t^2 +1−2t +2t^3 ))dt let decompose F(t)=((1−t^2 )/(t^4 +2t^3 +2t^2 −2t +1)) the roots of p(x)=t^4 +2t^3 +2t^2 −2t +1 are α(1+i) ,α(1−i) ,(α+1)(−1+i) and (α+1)(−1−i) with α=0,366 ⇒F(t)=((1−t^2 )/((t−α(1+i)(t−α(1−i))(t−(α+1)(−1+i))(t−(α+1)(−1−i)})) =((1−t^2 )/({t^2 −2Re(α(1+i))t +∣α(1+i)∣^2 }{t^2 −2Re(α+1)(−1+i)t +∣(α+1)(−1+i)∣^2 )) =((1−t^2 )/({t^2 −2αt +2α^2 ){t^2 +2 (α+1)t +2(α+1)^2 })) =((at +b)/(t^2 −2αt +2α^2 )) +((ct +d)/(t^2 +2(α+1)t +2(α+1)^2 )) rest to find the coefficient a ,b... any way we have I=[ln(x+(√(1+x^2 )))]_0 ^(∫_0 ^(π/2) ((cosθ)/(2−sin(2θ)))dθ) =ln(∫_0 ^(π/2) ((cosθ)/(2−sin(2θ)))dθ +(√(1+(∫_0 ^(π/2) ((cosθ)/(2−sin(2θ)))dθ)^2 )))

$${we}\:{havw}\:\int_{−\pi} ^{\pi} {costdt}\:=\mathrm{2}\int_{\mathrm{0}} ^{\pi} {cost}\:{dt}\:=\mathrm{2}\left[{sint}\right]_{\mathrm{0}} ^{\pi} =\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}\theta}{\mathrm{2}−{sin}\left(\mathrm{2}\theta\right)}{d}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\theta}{\mathrm{1}−{cos}\theta\:{sin}\theta}{d}\theta=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left\{\mathrm{1}−\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}{t}^{\mathrm{3}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{2}{t}\:+\mathrm{2}{t}^{\mathrm{3}} }{dt} \\ $$$${let}\:{decompose}\:\:{F}\left({t}\right)=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{1}} \\ $$$${the}\:{roots}\:{of}\:{p}\left({x}\right)={t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{1}\:{are} \\ $$$$\alpha\left(\mathrm{1}+{i}\right)\:,\alpha\left(\mathrm{1}−{i}\right)\:,\left(\alpha+\mathrm{1}\right)\left(−\mathrm{1}+{i}\right)\:\:{and}\:\left(\alpha+\mathrm{1}\right)\left(−\mathrm{1}−{i}\right) \\ $$$${with}\:\alpha=\mathrm{0},\mathrm{366}\:\Rightarrow{F}\left({t}\right)=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}−\alpha\left(\mathrm{1}+{i}\right)\left({t}−\alpha\left(\mathrm{1}−{i}\right)\right)\left({t}−\left(\alpha+\mathrm{1}\right)\left(−\mathrm{1}+{i}\right)\right)\left({t}−\left(\alpha+\mathrm{1}\right)\left(−\mathrm{1}−{i}\right)\right\}\right.} \\ $$$$=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left\{{t}^{\mathrm{2}} −\mathrm{2}{Re}\left(\alpha\left(\mathrm{1}+{i}\right)\right){t}\:+\mid\alpha\left(\mathrm{1}+{i}\right)\mid^{\mathrm{2}} \right\}\left\{{t}^{\mathrm{2}} \:−\mathrm{2}{Re}\left(\alpha+\mathrm{1}\right)\left(−\mathrm{1}+{i}\right){t}\:+\mid\left(\alpha+\mathrm{1}\right)\left(−\mathrm{1}+{i}\right)\mid^{\mathrm{2}} \right.} \\ $$$$=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left\{{t}^{\mathrm{2}} −\mathrm{2}\alpha{t}\:\:+\mathrm{2}\alpha^{\mathrm{2}} \right)\left\{{t}^{\mathrm{2}} \:+\mathrm{2}\:\left(\alpha+\mathrm{1}\right){t}\:\:+\mathrm{2}\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} \right\}} \\ $$$$=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} −\mathrm{2}\alpha{t}\:+\mathrm{2}\alpha^{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{2}\left(\alpha+\mathrm{1}\right){t}\:+\mathrm{2}\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${rest}\:{to}\:{find}\:{the}\:{coefficient}\:{a}\:,{b}…\:{any}\:{way}\:{we}\:{have} \\ $$$${I}=\left[{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cos}\theta}{\mathrm{2}−{sin}\left(\mathrm{2}\theta\right)}{d}\theta} \\ $$$$={ln}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cos}\theta}{\mathrm{2}−{sin}\left(\mathrm{2}\theta\right)}{d}\theta\:+\sqrt{\mathrm{1}+\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cos}\theta}{\mathrm{2}−{sin}\left(\mathrm{2}\theta\right)}{d}\theta\right)^{\mathrm{2}} }\right) \\ $$$$ \\ $$

Commented by TawaTawa last updated on 07/Feb/20