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Question Number 43490 by peter frank last updated on 11/Sep/18
evaluate ∫(√(tan𝛉 dΞΈ))
$$\boldsymbol{\mathrm{evaluate}} \\ $$$$\int\sqrt{\boldsymbol{\mathrm{tan}\theta}\:\boldsymbol{\mathrm{d}}\theta} \\ $$
Commented by maxmathsup by imad last updated on 11/Sep/18
let I =∫ (√(tanΞΈ))dΞΈ changement (√(tanΞΈ)) =x give tanΞΈ=x^2 β‡ŽΞΈ=arctan(x^2 ) β‡’ I = ∫ x ((2x)/(1+x^4 )) dx = ∫ ((2x^2 )/(1+x^4 )) dx let decompose F(x)=((2x^2 )/(1+x^4 )) F(x) = ((2x^2 )/((x^2 +1)^2 βˆ’2x^2 )) =((2x^2 )/((x^2 +(√2)x +1)(x^2 βˆ’(√2)x+1))) =((ax+b)/(x^2 +(√2)x+1)) +((cx+d)/(x^2 βˆ’(√2)x+1)) F(βˆ’x) =F(x) β‡’((βˆ’ax +b)/(x^2 βˆ’(√2)x+1)) +((βˆ’cx+d)/(x^2 +(√2)x +1)) =F(x) β‡’c=βˆ’a and b=d β‡’ F(x) = ((ax+b)/(x^2 +(√2)x +1)) +((βˆ’ax +b)/(x^2 βˆ’(√2)x +1)) F(0) =0 = 2b β‡’b=0 β‡’F(x)=((ax)/(x^2 +(√2)x+1)) βˆ’((ax)/(x^2 βˆ’(√2)x +1)) F(1) =1 = (a/(2+(√2))) βˆ’(a/(2βˆ’(√2))) β‡’(((βˆ’2(√2))/2))a=1 β‡’a=βˆ’(1/( (√2))) β‡’ F(x)= (1/( (√2))){ (1/(x^2 βˆ’(√2)x +1)) βˆ’(1/(x^2 +(√2)x +1))} β‡’ I =(1/( (√2))) ∫ (dx/(x^2 βˆ’(√2)x +1)) βˆ’(1/( (√2))) ∫ (dx/(x^2 +(√2)x+1)) but ∫ (dx/(x^2 βˆ’(√2)x +1)) =∫ (dx/(x^2 βˆ’(2/( (√2)))x +(1/2)+(1/2))) = ∫ (dx/((xβˆ’(1/( (√2))))^2 +(1/2))) =_(xβˆ’(1/( (√2))) =(u/( (√2)))) ∫ (1/((1/2)(1+u^2 ))) (du/( (√2))) = (√2) ∫ (du/(1+u^2 )) =(√2)arctan(u) +c_1 =(√2)arctan(x(√2)βˆ’1) +c_1 =(√2)arctan((√(2tanΞΈ))βˆ’1) +c_1 also we have ∫ (dx/(x^2 +(√2)x+1)) = ∫ (dx/((x+(1/( (√2))))^2 +(1/2))) =_(x+(1/( (√2)))=(u/( (√2)))) ∫ (1/((1/2)(1+u^2 ))) (du/( (√2))) =(√2)arctan((√2)x+1) =(√2)arctan((√(2tanΞΈ))+1) β‡’ ∫ (√(tanΞΈ))dΞΈ = arctan((√(2tanΞΈ))βˆ’1) βˆ’arctan((√(2tanΞΈ))+1) +c
$${let}\:{I}\:=\int\:\sqrt{{tan}\theta}{d}\theta\:\:\:\:{changement}\:\sqrt{{tan}\theta}\:={x}\:{give}\:{tan}\theta={x}^{\mathrm{2}} \:\nLeftrightarrow\theta={arctan}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${I}\:=\:\int\:\:{x}\:\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:=\:\int\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:βˆ’\mathrm{2}{x}^{\mathrm{2}} }\:=\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}\right)\left({x}^{\mathrm{2}} βˆ’\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)} \\ $$$$=\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} βˆ’\sqrt{\mathrm{2}}{x}+\mathrm{1}} \\ $$$${F}\left(βˆ’{x}\right)\:={F}\left({x}\right)\:\Rightarrow\frac{βˆ’{ax}\:+{b}}{{x}^{\mathrm{2}} \:βˆ’\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+\frac{βˆ’{cx}+{d}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:={F}\left({x}\right)\:\Rightarrow{c}=βˆ’{a}\:{and}\:{b}={d}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:+\frac{βˆ’{ax}\:+{b}}{{x}^{\mathrm{2}} \:βˆ’\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\:\mathrm{2}{b}\:\Rightarrow{b}=\mathrm{0}\:\Rightarrow{F}\left({x}\right)=\frac{{ax}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:βˆ’\frac{{ax}}{{x}^{\mathrm{2}} βˆ’\sqrt{\mathrm{2}}{x}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{1}\right)\:=\mathrm{1}\:=\:\:\frac{{a}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:βˆ’\frac{{a}}{\mathrm{2}βˆ’\sqrt{\mathrm{2}}}\:\Rightarrow\left(\frac{βˆ’\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right){a}=\mathrm{1}\:\Rightarrow{a}=βˆ’\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} βˆ’\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:βˆ’\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\right\}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} βˆ’\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:βˆ’\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:{but} \\ $$$$\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} βˆ’\sqrt{\mathrm{2}}{x}\:+\mathrm{1}}\:=\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:βˆ’\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}}{x}\:\:+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\int\:\:\:\:\frac{{dx}}{\left({x}βˆ’\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=_{{x}βˆ’\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:=\frac{{u}}{\:\sqrt{\mathrm{2}}}} \:\:\:\:\:\int\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\frac{{du}}{\:\sqrt{\mathrm{2}}}\:=\:\sqrt{\mathrm{2}}\:\:\int\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}}{arctan}\left({u}\right)\:+{c}_{\mathrm{1}} \\ $$$$=\sqrt{\mathrm{2}}{arctan}\left({x}\sqrt{\mathrm{2}}βˆ’\mathrm{1}\right)\:+{c}_{\mathrm{1}} =\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}{tan}\theta}βˆ’\mathrm{1}\right)\:+{c}_{\mathrm{1}} \:\:\:{also}\:{we}\:{have} \\ $$$$\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:=\:\int\:\:\:\:\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:=_{{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{{u}}{\:\sqrt{\mathrm{2}}}} \:\:\:\:\int\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\frac{{du}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:=\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}{tan}\theta}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\int\:\sqrt{{tan}\theta}{d}\theta\:=\:{arctan}\left(\sqrt{\mathrm{2}{tan}\theta}βˆ’\mathrm{1}\right)\:βˆ’{arctan}\left(\sqrt{\mathrm{2}{tan}\theta}+\mathrm{1}\right)\:+{c} \\ $$
Commented by peter frank last updated on 11/Sep/18
thanks
$${thanks}\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18
t^2 =tanΞΈ 2tdt=sec^2 ΞΈ dΞΈ ((2t)/(1+t^4 ))dt=dΞΈ ∫((2t)/(1+t^4 ))Γ—tdt ∫((2t^2 )/(1+t^4 ))dt ∫(2/(t^2 +(1/t^2 )))dt ∫((1βˆ’(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt ∫((1βˆ’(1/t^2 ))/((t+(1/t))^2 βˆ’((√2) )^2 ))dt+∫((1+(1/t^2 ))/((tβˆ’(1/t))^2 +((√2) )^2 ))dt ∫((d(t+(1/t)))/((t+(1/t))^2 βˆ’((√2) )^2 ))+∫((d(tβˆ’(1/t)))/((tβˆ’(1/t))^2 +((√2) )^2 )) (1/(2(√2)))ln∣(((t+(1/t))βˆ’(√2) )/((t+(1/t))+(√2) ))∣+(1/( (√2)))tan^(βˆ’1) (((tβˆ’(1/t))/( (√2) )))+c (1/(2(√2) ))ln∣(((√(tanΞΈ)) +(1/( (√(tanΞΈ)) ))βˆ’(√2))/( (√(tanΞΈ)) +(1/( (√(tanΞΈ)) ))+(√2)))∣+(1/( (√2) ))tan^(βˆ’1) ((((√(tanΞΈ)) βˆ’(1/( (√(tanΞΈ)) )))/( (√2) )))+c
$${t}^{\mathrm{2}} ={tan}\theta \\ $$$$\mathrm{2}{tdt}={sec}^{\mathrm{2}} \theta\:{d}\theta \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}={d}\theta \\ $$$$\int\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }Γ—{tdt} \\ $$$$\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\int\frac{\mathrm{2}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\int\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\int\frac{\mathrm{1}βˆ’\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} βˆ’\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{dt}+\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}βˆ’\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{dt} \\ $$$$\int\frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} βˆ’\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }+\int\frac{{d}\left({t}βˆ’\frac{\mathrm{1}}{{t}}\right)}{\left({t}βˆ’\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{\left({t}+\frac{\mathrm{1}}{{t}}\right)βˆ’\sqrt{\mathrm{2}}\:}{\left({t}+\frac{\mathrm{1}}{{t}}\right)+\sqrt{\mathrm{2}}\:}\mid+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{βˆ’\mathrm{1}} \left(\frac{{t}βˆ’\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}\:}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}{ln}\mid\frac{\sqrt{{tan}\theta}\:+\frac{\mathrm{1}}{\:\sqrt{{tan}\theta}\:}βˆ’\sqrt{\mathrm{2}}}{\:\sqrt{{tan}\theta}\:+\frac{\mathrm{1}}{\:\sqrt{{tan}\theta}\:}+\sqrt{\mathrm{2}}}\mid+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}{tan}^{βˆ’\mathrm{1}} \left(\frac{\sqrt{{tan}\theta}\:βˆ’\frac{\mathrm{1}}{\:\sqrt{{tan}\theta}\:}}{\:\sqrt{\mathrm{2}}\:}\right)+{c} \\ $$$$ \\ $$
Commented by peter frank last updated on 11/Sep/18
thanks
$${thanks} \\ $$

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