Question Number 43535 by maxmathsup by imad last updated on 11/Sep/18
$$\left.\mathrm{1}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\sqrt{\mathrm{1}+{tan}\theta}{d}\theta\:. \\ $$
Commented by MJS last updated on 12/Sep/18
$$\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{long}\:\mathrm{way} \\ $$$$\mathrm{start}\:\mathrm{with} \\ $$$$\sqrt{\mathrm{1}+\mathrm{tan}\:\theta}=\mathrm{sec}^{\mathrm{2}} \:\theta\:\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:\theta}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta} \\ $$$$\mathrm{then}\:{t}=\mathrm{1}+\mathrm{tan}\:\theta\:\rightarrow\:{d}\theta=\frac{{dt}}{\mathrm{sec}^{\mathrm{2}} \:\theta}\:\mathrm{which}\:\mathrm{leads} \\ $$$$\mathrm{to}\:\frac{\sqrt{{t}}}{\mathrm{1}+\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{then}\:{u}=\sqrt{{t}}\:\rightarrow\:{dt}=\mathrm{2}\sqrt{{t}}{du}\:\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{\mathrm{2}{v}^{\mathrm{2}} }{\mathrm{1}+\left({v}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{v}^{\mathrm{2}} }{{v}^{\mathrm{4}} −\mathrm{2}{v}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{factorise}\:\mathrm{the}\:\mathrm{denominator}\:\mathrm{and}\:\mathrm{decompose} \\ $$$$\mathrm{the}\:\mathrm{fraction}… \\ $$
Commented by maxmathsup by imad last updated on 12/Sep/18
![let A = ∫_(π/4) ^(π/3) (√(1+tanθ))dθ changement (√(1+tanθ))=x give 1+tanθ =x^2 ⇒ θ =artan(x^2 −1) ⇒dθ =((2x)/(1+(x^2 −1)^2 ))dx ⇒ A = ∫_(√2) ^(√(1+(√3))) x ((2xdx)/(1+(x^2 −1)^2 )) = ∫_(√2) ^(√(1+(√3))) ((2x^2 )/(1+x^4 −2x^2 +1))dx =∫_(√2) ^(√(1+(√3))) ((2x^2 )/(x^4 −2x^2 +2))dx let decompose inside C(x) F(x)=((2x^2 )/(x^4 −2x^2 +2)) roots of x^4 −2x^2 +2 →x^2 =t ⇒t^2 −2t +2 →Δ^′ =−1 ⇒ z_1 =1+i and z_2 =1−i ⇒F(x)=(a/(x−z_1 )) +(b/(x−z_2 )) a=lim_(x→z_1 ) (x−z_1 )F(x)=((2z_1 ^2 )/(z_1 −z_2 )) = ((2(1+i)^2 )/(2i)) =2 b =lim_(x→z_2 ) (x−z_2 )F(x)=((2z_2 ^2 )/(z_2 −z_1 )) = ((2(1−i)^2 )/(−2i)) =2 ⇒ F(x) =(2/(x^2 −1−i)) +(2/(x^2 −1+i)) ⇒ A = ∫_(√2) ^(√(1+(√3))) ((2dx)/(x^2 −1−i)) + ∫_(√2) ^(√(1+(√3))) ((2dx)/(x^2 −1+i)) =H +H^− = 2Re(H) = 2 Re( ∫_(√2) ^(√(1+(√3))) ((2dx)/(x^2 −1−i))) let decompose f(x) =(2/(x^2 −(1+i))) ⇒f(x)= (2/((x−(√(1+i)))(x+(√(1+i))))) wd have 1+i =(√2)e^((iπ)/4) ⇒(√(1+i))=2^(1/4) e^((iπ)/8) ⇒f(x) =(2/((x−2^(1/4) e^((iπ)/8) )(x+2^(1/4) e^((iπ)/8) ))) =(a/(x−2^(1/4) e^((iπ)/8) )) +(b/(x+2^(1/4) e^((iπ)/8) )) a = (2/(2 2^(1/4) e^((iπ)/8) )) =2^(−(1/4)) e^(−((iπ)/8)) and b = (2/(−2.2^(1/4) e^((iπ)/8) )) =−2^(1/4) e^(−((iπ)/8)) ⇒ f(x) =2^(1/4) e^(−((iπ)/8)) { (1/(x−2^(1/4) e^((iπ)/8) )) − (1/(x+2^(1/4) e^((iπ)/8) ))}⇒∫_(√2) ^(√(1+(√3))) ((2dx)/(x^2 −1−i)) =2^(1/4) e^(−((iπ)/8)) { ∫_(√2) ^(√(1+(√3))) (dx/(x−2^(1/4) e^((iπ)/8) )) −∫_(√2) ^(√(1+(√3))) (dx/(x+2^(1/4) e^((iπ)/8) ))} let find ∫ (dx/(x−z)) with z =a+ib ∫ (dx/(x−z)) = ∫ (dx/(x−a−ib)) = ∫ ((x−a +ib)/((x−a)^2 +b^2 )) dx =(1/2)ln{(x−a)^2 +b^2 } +ib ∫ (dx/((x−a)^2 +b^2 )) but ∫ (dx/((x−a)^2 +b^2 )) =_(x−a=bt) ∫ ((bdt)/(b^2 (1+t^2 ))) =(1/b) arctan(((x−a)/b)) ⇒ ∫ (dx/(x−z)) =(1/2)ln{(x−a)^2 +b^2 } +i arctan(((x−a)/b))we have 2^(1/4) e^((iπ)/8) =2^(1/4) cos((π/8)) +i 2^(1/4) sin((π/8)) =a+ib ⇒ ∫_(√2) ^(√(1+(√3))) (dx/(x−2^(1/4) e^((iπ)/8) )) =[(1/2)ln{(x−2^(1/4) cos((π/8)))^2 +(√2)sin^2 ((π/8))}]_(√2) ^(√(1+(√3))) +i[ arctan{((x−2^(1/4) cos((π/8)))/(2^(1/4) sin((π/8))))}]_(√2) ^(√(1+(√3))) so the value of A is known after calculus..](https://www.tinkutara.com/question/Q43609.png)
$${let}\:{A}\:=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\sqrt{\mathrm{1}+{tan}\theta}{d}\theta\:{changement}\:\sqrt{\mathrm{1}+{tan}\theta}={x}\:{give}\:\mathrm{1}+{tan}\theta\:={x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\theta\:={artan}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:\Rightarrow{d}\theta\:=\frac{\mathrm{2}{x}}{\mathrm{1}+\left({x}^{\mathrm{2}} \:−\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${A}\:=\:\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} {x}\:\frac{\mathrm{2}{xdx}}{\mathrm{1}+\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:=\:\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} \:−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}}{dx}\:{let}\:{decompose}\:{inside}\:{C}\left({x}\right)\:{F}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}} \\ $$$${roots}\:{of}\:{x}^{\mathrm{4}} \:−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}\:\rightarrow{x}^{\mathrm{2}} ={t}\:\Rightarrow{t}^{\mathrm{2}} \:−\mathrm{2}{t}\:+\mathrm{2}\:\rightarrow\Delta^{'} =−\mathrm{1}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\mathrm{1}+{i}\:{and}\:{z}_{\mathrm{2}} =\mathrm{1}−{i}\:\Rightarrow{F}\left({x}\right)=\frac{{a}}{{x}−{z}_{\mathrm{1}} }\:+\frac{{b}}{{x}−{z}_{\mathrm{2}} } \\ $$$${a}={lim}_{{x}\rightarrow{z}_{\mathrm{1}} } \left({x}−{z}_{\mathrm{1}} \right){F}\left({x}\right)=\frac{\mathrm{2}{z}_{\mathrm{1}} ^{\mathrm{2}} }{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }\:=\:\frac{\mathrm{2}\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }{\mathrm{2}{i}}\:=\mathrm{2} \\ $$$${b}\:={lim}_{{x}\rightarrow{z}_{\mathrm{2}} } \left({x}−{z}_{\mathrm{2}} \right){F}\left({x}\right)=\frac{\mathrm{2}{z}_{\mathrm{2}} ^{\mathrm{2}} }{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }\:=\:\frac{\mathrm{2}\left(\mathrm{1}−{i}\right)^{\mathrm{2}} }{−\mathrm{2}{i}}\:=\mathrm{2}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}−{i}}\:+\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}+{i}}\:\Rightarrow\:{A}\:=\:\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}} −\mathrm{1}−{i}}\:+\:\:\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\:\:\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}} −\mathrm{1}+{i}} \\ $$$$={H}\:+\overset{−} {{H}}\:=\:\mathrm{2}{Re}\left({H}\right)\:=\:\mathrm{2}\:{Re}\left(\:\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\:\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}} −\mathrm{1}−{i}}\right)\:\:{let}\:{decompose} \\ $$$${f}\left({x}\right)\:=\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\left(\mathrm{1}+{i}\right)}\:\Rightarrow{f}\left({x}\right)=\:\frac{\mathrm{2}}{\left({x}−\sqrt{\mathrm{1}+{i}}\right)\left({x}+\sqrt{\mathrm{1}+{i}}\right)}\:\:{wd}\:{have}\: \\ $$$$\mathrm{1}+{i}\:=\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\Rightarrow\sqrt{\mathrm{1}+{i}}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} \:\Rightarrow{f}\left({x}\right)\:=\frac{\mathrm{2}}{\left({x}−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)\left({x}+\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} \right)} \\ $$$$=\frac{{a}}{{x}−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} }\:+\frac{{b}}{{x}+\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} } \\ $$$${a}\:=\:\frac{\mathrm{2}}{\mathrm{2}\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} }\:=\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \:\:{and}\:{b}\:=\:\frac{\mathrm{2}}{−\mathrm{2}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} }\:=−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \:\left\{\:\:\frac{\mathrm{1}}{{x}−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} }\:−\:\frac{\mathrm{1}}{{x}+\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} }\right\}\Rightarrow\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}} −\mathrm{1}−{i}} \\ $$$$=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}\pi}{\mathrm{8}}} \left\{\:\:\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\:\frac{{dx}}{{x}−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} }\:−\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\:\frac{{dx}}{{x}+\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} }\right\}\:{let}\:{find} \\ $$$$\int\:\:\:\frac{{dx}}{{x}−{z}}\:\:\:{with}\:{z}\:={a}+{ib} \\ $$$$\int\:\:\:\frac{{dx}}{{x}−{z}}\:=\:\int\:\:\:\frac{{dx}}{{x}−{a}−{ib}}\:=\:\int\:\:\:\frac{{x}−{a}\:+{ib}}{\left({x}−{a}\right)^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\left({x}−{a}\right)^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right\}\:+{ib}\:\int\:\:\:\frac{{dx}}{\left({x}−{a}\right)^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:\:{but}\:\: \\ $$$$\int\:\:\:\frac{{dx}}{\left({x}−{a}\right)^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:=_{{x}−{a}={bt}} \:\int\:\:\:\:\frac{{bdt}}{{b}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{b}}\:{arctan}\left(\frac{{x}−{a}}{{b}}\right)\:\Rightarrow \\ $$$$\int\:\:\:\frac{{dx}}{{x}−{z}}\:\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\left({x}−{a}\right)^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right\}\:+{i}\:{arctan}\left(\frac{{x}−{a}}{{b}}\right){we}\:{have} \\ $$$$\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} \:=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{cos}\left(\frac{\pi}{\mathrm{8}}\right)\:+{i}\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{sin}\left(\frac{\pi}{\mathrm{8}}\right)\:={a}+{ib}\:\Rightarrow \\ $$$$\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\frac{{dx}}{{x}−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}\pi}{\mathrm{8}}} }\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\left({x}−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{cos}\left(\frac{\pi}{\mathrm{8}}\right)\right)^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right\}\right]_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \\ $$$$+{i}\left[\:{arctan}\left\{\frac{{x}−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{cos}\left(\frac{\pi}{\mathrm{8}}\right)}{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{sin}\left(\frac{\pi}{\mathrm{8}}\right)}\right\}\right]_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{3}}}} \:\:\:{so}\:{the}\:{value}\:{of}\:\:{A}\:{is}\:{known}\:{after}\: \\ $$$${calculus}.. \\ $$$$ \\ $$
Answered by behi83417@gmail.com last updated on 12/Sep/18
![tgθ=tg^2 x⇒(1+tg^2 θ)dθ=2tgx(1+tg^2 x)dx I=∫ (1/(cosx)).((2tgx(1+tg^2 x))/((1+tg^4 x)))dx= =∫ ((((2sinx)/(cosx)).(1/(cos^2 x)))/(cosx(((sin^4 x+cos^4 x)/(cos^4 x)))))dx= =∫ ((2sinx)/(sin^4 x+cos^4 x))dx= =∫((2sinx)/(1−2cos^2 x(1−cos^2 x)))dx= =∫((2sinx)/(2cos^4 x−2cos^2 x+1))dx=−∫(du/(2u^4 −2u^2 +1))... 2u^4 −2u^2 +1=0⇒u^2 =((2±(√(4−8)))/4)= =(1/2)(1±i)=α^2 ,β^2 (α,β∈C) [α^2 +β^2 =1,α^2 −β^2 =i,αβ=(1/( (√2)))] (α+β)^2 =1+2×(1/( (√2)))=(√2)+1⇒α+β=(√((√2)+1)) α−β=−((√2)−1)⇒α−β=i(√((√2)−1)) ⇒ { ((α=(1/2)[(√((√2)+1))+i(√((√2)−1))))),((β=(1/2)[(√((√2)+1))−i(√((√2)−1))))) :} I=∫(du/((u−α)(u+α)(u−β)(u+β)))= =(i/(2α)).ln((u+α)/(u−α))−(i/(2β)).ln((u+β)/(u−β))+const.= =(i/(2α))ln((u+α)/(u−β))−(i/(2β))ln((u+β)/(u−β))+const.== =(i/(2α))ln(((√(2+tgθ))+α)/( (√(2+tgθ))−α))−(i/(2β))ln(((√(2+tgθ))+β)/( (√(2+tgθ))−β))+const [(1/(Π(u−α)))=Σ(k/((u−α))) u=α⇒k_1 =(1/((2α)(α^2 −β^2 )))=((−i)/(2α)) u=−α⇒k_2 =(1/((2α)(β^2 −α^2 )))=(i/(2α)) u=β⇒k_3 =(1/(2β(β^2 −α^2 )))=(i/(2β)) u=−β⇒k_4 =(1/(2β(α^2 −β^2 )))=((−i)/(2β))]](https://www.tinkutara.com/question/Q43561.png)
$${tg}\theta={tg}^{\mathrm{2}} {x}\Rightarrow\left(\mathrm{1}+{tg}^{\mathrm{2}} \theta\right){d}\theta=\mathrm{2}{tgx}\left(\mathrm{1}+{tg}^{\mathrm{2}} {x}\right){dx} \\ $$$${I}=\int\:\:\:\frac{\mathrm{1}}{{cosx}}.\frac{\mathrm{2}{tgx}\left(\mathrm{1}+{tg}^{\mathrm{2}} {x}\right)}{\left(\mathrm{1}+{tg}^{\mathrm{4}} {x}\right)}{dx}= \\ $$$$=\int\:\:\frac{\frac{\mathrm{2}{sinx}}{{cosx}}.\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}}{{cosx}\left(\frac{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}}{{cos}^{\mathrm{4}} {x}}\right)}{dx}= \\ $$$$=\int\:\:\frac{\mathrm{2}{sinx}}{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}}{dx}= \\ $$$$=\int\frac{\mathrm{2}{sinx}}{\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} {x}\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)}{dx}= \\ $$$$=\int\frac{\mathrm{2}{sinx}}{\mathrm{2}{cos}^{\mathrm{4}} {x}−\mathrm{2}{cos}^{\mathrm{2}} {x}+\mathrm{1}}{dx}=−\int\frac{{du}}{\mathrm{2}{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{1}}… \\ $$$$ \\ $$$$\mathrm{2}{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\Rightarrow{u}^{\mathrm{2}} =\frac{\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{8}}}{\mathrm{4}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm{i}\right)=\alpha^{\mathrm{2}} ,\beta^{\mathrm{2}} \:\:\:\left(\alpha,\beta\in\boldsymbol{\mathrm{C}}\right) \\ $$$$\left[\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\mathrm{1},\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} ={i},\alpha\beta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right] \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}+\mathrm{1}\Rightarrow\alpha+\beta=\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}} \\ $$$$\alpha−\beta=−\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\Rightarrow\alpha−\beta={i}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$\Rightarrow\begin{cases}{\alpha=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}+{i}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)}\\{\beta=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}−{i}\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)}\end{cases} \\ $$$${I}=\int\frac{{du}}{\left({u}−\alpha\right)\left({u}+\alpha\right)\left({u}−\beta\right)\left({u}+\beta\right)}= \\ $$$$=\frac{{i}}{\mathrm{2}\alpha}.{ln}\frac{{u}+\alpha}{{u}−\alpha}−\frac{{i}}{\mathrm{2}\beta}.{ln}\frac{{u}+\beta}{{u}−\beta}+{const}.= \\ $$$$=\frac{{i}}{\mathrm{2}\alpha}{ln}\frac{{u}+\alpha}{{u}−\beta}−\frac{{i}}{\mathrm{2}\beta}{ln}\frac{{u}+\beta}{{u}−\beta}+{const}.== \\ $$$$=\frac{{i}}{\mathrm{2}\alpha}{ln}\frac{\sqrt{\mathrm{2}+{tg}\theta}+\alpha}{\:\sqrt{\mathrm{2}+{tg}\theta}−\alpha}−\frac{{i}}{\mathrm{2}\beta}{ln}\frac{\sqrt{\mathrm{2}+{tg}\theta}+\beta}{\:\sqrt{\mathrm{2}+{tg}\theta}−\beta}+{const} \\ $$$$ \\ $$$$\left[\frac{\mathrm{1}}{\Pi\left({u}−\alpha\right)}=\Sigma\frac{{k}}{\left({u}−\alpha\right)}\right. \\ $$$${u}=\alpha\Rightarrow{k}_{\mathrm{1}} =\frac{\mathrm{1}}{\left(\mathrm{2}\alpha\right)\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)}=\frac{−{i}}{\mathrm{2}\alpha} \\ $$$${u}=−\alpha\Rightarrow{k}_{\mathrm{2}} =\frac{\mathrm{1}}{\left(\mathrm{2}\alpha\right)\left(\beta^{\mathrm{2}} −\alpha^{\mathrm{2}} \right)}=\frac{{i}}{\mathrm{2}\alpha} \\ $$$${u}=\beta\Rightarrow{k}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}\beta\left(\beta^{\mathrm{2}} −\alpha^{\mathrm{2}} \right)}=\frac{{i}}{\mathrm{2}\beta} \\ $$$$\left.{u}=−\beta\Rightarrow{k}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}\beta\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)}=\frac{−{i}}{\mathrm{2}\beta}\right] \\ $$
Answered by MJS last updated on 13/Sep/18
![∫(√(1+tan θ)) dθ=∫((sec^2 θ)/(1+tan^2 θ))(√(1+tan θ)) dθ= [t=1+tan θ → dθ=(dt/(sec^2 θ))] =∫((√t)/(1+(t−1)^2 ))dt= [u=(√t) → dt=2(√t)du] =2∫(u^2 /(1+(u^2 −1)^2 ))du=2∫(u^2 /(u^4 −2u^2 +2))du= =2∫(u^2 /((u^2 −(√(2+(√8)))u+(√2))(u^2 +(√(2+(√8)))u+(√2))))du= [a=(√(2+(√8))); b=(√2)] =2∫(u^2 /((u^2 −au+b)(u^2 +au+b)))du= =2∫((u/(2a(u^2 −au+b)))−(u/(2a(u^2 +au+b))))du= =(1/a)∫(u/(u^2 −au+b))du−(1/a)∫(u/(u^2 +au+b))du ∫(u/(u^2 −au+b))du=∫(((2u−a)/(2(u^2 −au+b)))+(a/(2(u^2 −au+b)))du= =(1/2)∫((2u−a)/(u^2 −au+b))du+(a/2)∫(du/(u^2 −au+b))= [these are standard integrals] =(1/2)ln (u^2 −au+b) +(a/( (√(4b−a^2 ))))arctan ((2u−a)/( (√(4b−a^2 )))) ∫(u/(u^2 +au+b))du=∫(((2u+a)/(2(u^2 +au+b)))−(a/(2(u^2 +au+b))))du= =(1/2)∫((2u+a)/(u^2 +au+b))du−(a/2)∫(du/(u^2 +au+b))= [again standard integrals] =(1/2)ln (u^2 +au+b) −(a/( (√(4b−a^2 ))))arctan ((2u+a)/( (√(4b−a^2 )))) =(1/(2a))(ln (u^2 −au+b) +ln (u^2 +au+b))+(1/( (√(4b−a^2 ))))(arctan ((2u−a)/( (√(4b−a^2 )))) −arctan ((2u+a)/( (√(4b−a^2 )))))= =(1/(2a))ln (u^4 +(2b−a^2 )u^2 +b^2 ) +(1/( (√(4b−a^2 ))))(arctan ((2u−a)/( (√(4b−a^2 )))) −arctan ((2u+a)/( (√(4b−a^2 )))))= =(1/(2a))ln ∣t^2 +(2b−a^2 )t+b^2 ∣ +(1/( (√(4b−a^2 ))))(arctan ((2(√t)−a)/( (√(4b−a^2 )))) −arctan ((2(√t)+a)/( (√(4b−a^2 )))))+C sorry no time to finish t=1+tan θ a=(√(2+(√8))) b=(√2)](https://www.tinkutara.com/question/Q43673.png)
$$\int\sqrt{\mathrm{1}+\mathrm{tan}\:\theta}\:{d}\theta=\int\frac{\mathrm{sec}^{\mathrm{2}} \:\theta}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}\sqrt{\mathrm{1}+\mathrm{tan}\:\theta}\:{d}\theta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{1}+\mathrm{tan}\:\theta\:\rightarrow\:{d}\theta=\frac{{dt}}{\mathrm{sec}^{\mathrm{2}} \:\theta}\right] \\ $$$$=\int\frac{\sqrt{{t}}}{\mathrm{1}+\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\sqrt{{t}}\:\rightarrow\:{dt}=\mathrm{2}\sqrt{{t}}{du}\right] \\ $$$$=\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{\mathrm{1}+\left({u}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{du}=\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}}{du}= \\ $$$$=\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} −\sqrt{\mathrm{2}+\sqrt{\mathrm{8}}}{u}+\sqrt{\mathrm{2}}\right)\left({u}^{\mathrm{2}} +\sqrt{\mathrm{2}+\sqrt{\mathrm{8}}}{u}+\sqrt{\mathrm{2}}\right)}{du}= \\ $$$$\:\:\:\:\:\left[{a}=\sqrt{\mathrm{2}+\sqrt{\mathrm{8}}};\:{b}=\sqrt{\mathrm{2}}\right] \\ $$$$=\mathrm{2}\int\frac{{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} −{au}+{b}\right)\left({u}^{\mathrm{2}} +{au}+{b}\right)}{du}= \\ $$$$=\mathrm{2}\int\left(\frac{{u}}{\mathrm{2}{a}\left({u}^{\mathrm{2}} −{au}+{b}\right)}−\frac{{u}}{\mathrm{2}{a}\left({u}^{\mathrm{2}} +{au}+{b}\right)}\right){du}= \\ $$$$=\frac{\mathrm{1}}{{a}}\int\frac{{u}}{{u}^{\mathrm{2}} −{au}+{b}}{du}−\frac{\mathrm{1}}{{a}}\int\frac{{u}}{{u}^{\mathrm{2}} +{au}+{b}}{du} \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{{u}}{{u}^{\mathrm{2}} −{au}+{b}}{du}=\int\left(\frac{\mathrm{2}{u}−{a}}{\mathrm{2}\left({u}^{\mathrm{2}} −{au}+{b}\right)}+\frac{{a}}{\mathrm{2}\left({u}^{\mathrm{2}} −{au}+{b}\right.}\right){du}= \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{u}−{a}}{{u}^{\mathrm{2}} −{au}+{b}}{du}+\frac{{a}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} −{au}+{b}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{these}\:\mathrm{are}\:\mathrm{standard}\:\mathrm{integrals}\right] \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({u}^{\mathrm{2}} −{au}+{b}\right)\:+\frac{{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\mathrm{arctan}\:\frac{\mathrm{2}{u}−{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }} \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{{u}}{{u}^{\mathrm{2}} +{au}+{b}}{du}=\int\left(\frac{\mathrm{2}{u}+{a}}{\mathrm{2}\left({u}^{\mathrm{2}} +{au}+{b}\right)}−\frac{{a}}{\mathrm{2}\left({u}^{\mathrm{2}} +{au}+{b}\right)}\right){du}= \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{u}+{a}}{{u}^{\mathrm{2}} +{au}+{b}}{du}−\frac{{a}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} +{au}+{b}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{again}\:\mathrm{standard}\:\mathrm{integrals}\right] \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({u}^{\mathrm{2}} +{au}+{b}\right)\:−\frac{{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\mathrm{arctan}\:\frac{\mathrm{2}{u}+{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }} \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}}\left(\mathrm{ln}\:\left({u}^{\mathrm{2}} −{au}+{b}\right)\:+\mathrm{ln}\:\left({u}^{\mathrm{2}} +{au}+{b}\right)\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\left(\mathrm{arctan}\:\frac{\mathrm{2}{u}−{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\:−\mathrm{arctan}\:\frac{\mathrm{2}{u}+{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}}\mathrm{ln}\:\left({u}^{\mathrm{4}} +\left(\mathrm{2}{b}−{a}^{\mathrm{2}} \right){u}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\left(\mathrm{arctan}\:\frac{\mathrm{2}{u}−{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\:−\mathrm{arctan}\:\frac{\mathrm{2}{u}+{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{a}}\mathrm{ln}\:\mid{t}^{\mathrm{2}} +\left(\mathrm{2}{b}−{a}^{\mathrm{2}} \right){t}+{b}^{\mathrm{2}} \mid\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\left(\mathrm{arctan}\:\frac{\mathrm{2}\sqrt{{t}}−{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\:−\mathrm{arctan}\:\frac{\mathrm{2}\sqrt{{t}}+{a}}{\:\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}\right)+{C} \\ $$$$\mathrm{sorry}\:\mathrm{no}\:\mathrm{time}\:\mathrm{to}\:\mathrm{finish} \\ $$$${t}=\mathrm{1}+\mathrm{tan}\:\theta \\ $$$${a}=\sqrt{\mathrm{2}+\sqrt{\mathrm{8}}} \\ $$$${b}=\sqrt{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 13/Sep/18
$${thank}\:{you}\:{sir}. \\ $$