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Question Number 38074 by ajfour last updated on 21/Jun/18
∫(dx/(a+btan^2 x)) = ?
$$\int\frac{{dx}}{{a}+{b}\mathrm{tan}\:^{\mathrm{2}} {x}}\:=\:? \\ $$
Commented by prof Abdo imad last updated on 21/Jun/18
changement tanx=t give I = ∫ (1/(a+bt^2 )) (dt/(1+t^2 )) =∫ (dt/((1+t^2 )(a+bt^2 ))) let decompose F(t)= (1/((1+t^2 )(a+bt^2 ))) F(t)= ((αt +β)/(t^2 +1)) +((ct +d)/(bt^2 +a)) F(−t)=F(t) ⇒((−αt +β)/(t^2 +1)) +((−ct +d)/(bt^2 +a)) =F(t)⇒ α=c=0 ⇒F(t)=(β/(1+t^2 )) +(d/(bt^2 +a)) lim_(t→+∞) t^2 F(t)=0=β +(d/b) ⇒bβ +d=0 ⇒ d=−bβ ⇒F(t)=(β/(1+t^2 )) −((bβ)/(bt^2 +a)) F(0)= (1/a) =β −((bβ)/a) ⇒1=aβ −bβ ⇒β= (1/(a−b)) if a≠b ⇒F(t)= (1/(a−b)){ (1/(1+t^2 )) −(b/(bt^2 +a))} I = ∫ F(t)dt ⇒(a−b)I= ∫ (dt/(1+t^2 )) −∫ (dt/(t^2 +(a/b))) case1 (a/b)>0 we do the chang.t=(√(a/b)) u ∫ (dt/(t^2 +(a/b))) = ∫ (1/((a/b)(1+u^2 ))) (√(a/b)) du =(b/a) (√(a/b)) arctanu =(√(b/a)) arctan((√(b/a))tanx)⇒ (a−b)I= x −(√(b/a)) arctan((√(b/a))tanx)⇒ I= (1/(a−b)){ x−(√(b/a))arctan((√(b/a))tanx) +λ case2 (a/b)<0 ∫ (dt/(t^2 +(a/b))) =∫ (dt/(t^2 −((√(−(a/b))))^2 )) = (1/(2(√(−(a/b)))))∫ ( (1/(t−(√(−(a/b))))) −(1/(t +(√(−(a/b))))))dt = (1/(2(√(−(a/b)))))ln∣ ((t−(√(−(a/b))))/(t+(√(−(a/b)))))∣ ⇒ I = (1/(a−b)){ x −(1/(2(√(−(a/b))))) ln∣((tan(x) −(√(−(a/b))))/(tanx +(√(−(a/b)))))∣} +λ
$${changement}\:{tanx}={t}\:{give} \\ $$$${I}\:=\:\int\:\:\frac{\mathrm{1}}{{a}+{bt}^{\mathrm{2}} }\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+{bt}^{\mathrm{2}} \right)} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+{bt}^{\mathrm{2}} \right)} \\ $$$${F}\left({t}\right)=\:\frac{\alpha{t}\:+\beta}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}\:+{d}}{{bt}^{\mathrm{2}} \:+{a}} \\ $$$${F}\left(−{t}\right)={F}\left({t}\right)\:\Rightarrow\frac{−\alpha{t}\:+\beta}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{ct}\:+{d}}{{bt}^{\mathrm{2}} \:+{a}}\:={F}\left({t}\right)\Rightarrow \\ $$$$\alpha={c}=\mathrm{0}\:\Rightarrow{F}\left({t}\right)=\frac{\beta}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{{d}}{{bt}^{\mathrm{2}} \:+{a}} \\ $$$${lim}_{{t}\rightarrow+\infty} {t}^{\mathrm{2}} {F}\left({t}\right)=\mathrm{0}=\beta\:+\frac{{d}}{{b}}\:\Rightarrow{b}\beta\:+{d}=\mathrm{0}\:\Rightarrow \\ $$$${d}=−{b}\beta\:\Rightarrow{F}\left({t}\right)=\frac{\beta}{\mathrm{1}+{t}^{\mathrm{2}} }\:−\frac{{b}\beta}{{bt}^{\mathrm{2}} \:+{a}} \\ $$$${F}\left(\mathrm{0}\right)=\:\frac{\mathrm{1}}{{a}}\:=\beta\:−\frac{{b}\beta}{{a}}\:\Rightarrow\mathrm{1}={a}\beta\:−{b}\beta\:\Rightarrow\beta=\:\frac{\mathrm{1}}{{a}−{b}} \\ $$$${if}\:{a}\neq{b}\:\Rightarrow{F}\left({t}\right)=\:\frac{\mathrm{1}}{{a}−{b}}\left\{\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:−\frac{{b}}{{bt}^{\mathrm{2}} \:+{a}}\right\} \\ $$$${I}\:=\:\int\:{F}\left({t}\right){dt}\:\Rightarrow\left({a}−{b}\right){I}=\:\int\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:−\int\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{{a}}{{b}}} \\ $$$${case}\mathrm{1}\:\frac{{a}}{{b}}>\mathrm{0}\:\:{we}\:{do}\:{the}\:{chang}.{t}=\sqrt{\frac{{a}}{{b}}}\:{u} \\ $$$$\int\:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{{a}}{{b}}}\:=\:\int\:\:\:\:\frac{\mathrm{1}}{\frac{{a}}{{b}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\sqrt{\frac{{a}}{{b}}}\:{du} \\ $$$$=\frac{{b}}{{a}}\:\sqrt{\frac{{a}}{{b}}}\:{arctanu}\:=\sqrt{\frac{{b}}{{a}}}\:{arctan}\left(\sqrt{\frac{{b}}{{a}}}{tanx}\right)\Rightarrow \\ $$$$\left({a}−{b}\right){I}=\:{x}\:−\sqrt{\frac{{b}}{{a}}}\:{arctan}\left(\sqrt{\frac{{b}}{{a}}}{tanx}\right)\Rightarrow \\ $$$${I}=\:\frac{\mathrm{1}}{{a}−{b}}\left\{\:{x}−\sqrt{\frac{{b}}{{a}}}{arctan}\left(\sqrt{\frac{{b}}{{a}}}{tanx}\right)\:+\lambda\right. \\ $$$${case}\mathrm{2}\:\:\frac{{a}}{{b}}<\mathrm{0}\:\:\: \\ $$$$\int\:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{{a}}{{b}}}\:=\int\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} −\left(\sqrt{−\frac{{a}}{{b}}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−\frac{{a}}{{b}}}}\int\:\:\left(\:\:\:\frac{\mathrm{1}}{{t}−\sqrt{−\frac{{a}}{{b}}}}\:−\frac{\mathrm{1}}{{t}\:+\sqrt{−\frac{{a}}{{b}}}}\right){dt} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−\frac{{a}}{{b}}}}{ln}\mid\:\:\frac{{t}−\sqrt{−\frac{{a}}{{b}}}}{{t}+\sqrt{−\frac{{a}}{{b}}}}\mid\:\Rightarrow \\ $$$${I}\:=\:\frac{\mathrm{1}}{{a}−{b}}\left\{\:{x}\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{−\frac{{a}}{{b}}}}\:{ln}\mid\frac{{tan}\left({x}\right)\:−\sqrt{−\frac{{a}}{{b}}}}{{tanx}\:+\sqrt{−\frac{{a}}{{b}}}}\mid\right\}\:+\lambda \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 21/Jun/18
if a=b I = (1/a) ∫ (dx/(1+tan^2 x)) (a≠0) =(1/a) ∫ cos^2 x dx=(1/a) ∫ ((1+cos(2x))/2)dx = (x/(2a)) +(1/(4a)) sin(2x) + λ .
$${if}\:{a}={b}\:\:{I}\:\:=\:\frac{\mathrm{1}}{{a}}\:\int\:\:\:\:\:\frac{{dx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}\:\:\:\:\left({a}\neq\mathrm{0}\right) \\ $$$$=\frac{\mathrm{1}}{{a}}\:\int\:{cos}^{\mathrm{2}} {x}\:{dx}=\frac{\mathrm{1}}{{a}}\:\int\:\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{dx} \\ $$$$=\:\frac{{x}}{\mathrm{2}{a}}\:\:+\frac{\mathrm{1}}{\mathrm{4}{a}}\:{sin}\left(\mathrm{2}{x}\right)\:+\:\lambda\:. \\ $$
Commented by ajfour last updated on 21/Jun/18
Wow Sir, thank you .
$${Wow}\:{Sir},\:{thank}\:{you}\:. \\ $$
Commented by math khazana by abdo last updated on 21/Jun/18
nevermind sir Ajfour.
$${nevermind}\:{sir}\:{Ajfour}. \\ $$
Answered by behi83417@gmail.com last updated on 21/Jun/18
tg^2 x=(a/b)t⇒tgx=(√((a/b)t)) 2tgx(1+tg^2 x)dx=(a/b)dt⇒ 2(√((at)/b))(1+((at)/b))dx=(a/b)dt⇒dx=((adt)/(2(√((at)/b)).(b+at))) ⇒I=(1/(2(√(a/b))))∫(((adt)/( (√t)(b+at)))/(a+at))=m∫(dt/( (√t)(1+t)(b+at))) m=(2(√(a/b)))^(−1) ,t=u^2 ⇒dt=2udu I=m∫((2udu)/(u(1+u^2 )(b+au^2 )))=2m∫(du/((1+u^2 )(b+au^2 )))= =2m[a.(1/(4(√(ab))))tg^(−1) (a.(u/( (√(ab)))))−(1/4)tg^(−1) u]+const =(1/2)m[(√(a/b)).tg^(−1) ((√(a/b))u)−tg^(−1) u+const= =(1/2).(1/2)(√(b/a)).(√(a/b))tg^(−1) ((√(a/b)).(√(b/a)).tgx)− −(1/4).(√(b/a)).tg^(−1) ((√(b/a)).tgx)+const= =(1/4)x−(1/4).(√(b/a)).arctg[(√(b/a)).tgx]+const. ■
$${tg}^{\mathrm{2}} {x}=\frac{{a}}{{b}}{t}\Rightarrow{tgx}=\sqrt{\frac{{a}}{{b}}{t}} \\ $$$$\mathrm{2}{tgx}\left(\mathrm{1}+{tg}^{\mathrm{2}} {x}\right){dx}=\frac{{a}}{{b}}{dt}\Rightarrow \\ $$$$\mathrm{2}\sqrt{\frac{{at}}{{b}}}\left(\mathrm{1}+\frac{{at}}{{b}}\right){dx}=\frac{{a}}{{b}}{dt}\Rightarrow{dx}=\frac{{adt}}{\mathrm{2}\sqrt{\frac{{at}}{{b}}}.\left({b}+{at}\right)} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\frac{{a}}{{b}}}}\int\frac{\frac{{adt}}{\:\sqrt{{t}}\left({b}+{at}\right)}}{{a}+{at}}={m}\int\frac{{dt}}{\:\sqrt{{t}}\left(\mathrm{1}+{t}\right)\left({b}+{at}\right)} \\ $$$${m}=\left(\mathrm{2}\sqrt{\frac{{a}}{{b}}}\right)^{−\mathrm{1}} ,{t}={u}^{\mathrm{2}} \Rightarrow{dt}=\mathrm{2}{udu} \\ $$$${I}={m}\int\frac{\mathrm{2}{udu}}{{u}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({b}+{au}^{\mathrm{2}} \right)}=\mathrm{2}{m}\int\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({b}+{au}^{\mathrm{2}} \right)}= \\ $$$$=\mathrm{2}{m}\left[{a}.\frac{\mathrm{1}}{\mathrm{4}\sqrt{{ab}}}{tg}^{−\mathrm{1}} \left({a}.\frac{{u}}{\:\sqrt{{ab}}}\right)−\frac{\mathrm{1}}{\mathrm{4}}{tg}^{−\mathrm{1}} {u}\right]+{const} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{m}\left[\sqrt{\frac{{a}}{{b}}}.{tg}^{−\mathrm{1}} \left(\sqrt{\frac{{a}}{{b}}}{u}\right)−{tg}^{−\mathrm{1}} {u}+{const}=\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{{b}}{{a}}}.\sqrt{\frac{{a}}{{b}}}{tg}^{−\mathrm{1}} \left(\sqrt{\frac{{a}}{{b}}}.\sqrt{\frac{{b}}{{a}}}.{tgx}\right)− \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}.\sqrt{\frac{{b}}{{a}}}.{tg}^{−\mathrm{1}} \left(\sqrt{\frac{{b}}{{a}}}.{tgx}\right)+{const}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\boldsymbol{{x}}−\frac{\mathrm{1}}{\mathrm{4}}.\sqrt{\frac{\boldsymbol{{b}}}{\boldsymbol{{a}}}}.\boldsymbol{{arctg}}\left[\sqrt{\frac{\boldsymbol{{b}}}{\boldsymbol{{a}}}}.\boldsymbol{{tgx}}\right]+\boldsymbol{{const}}.\:\blacksquare \\ $$
Commented by ajfour last updated on 21/Jun/18
God bless you Sir.
$${God}\:{bless}\:{you}\:{Sir}. \\ $$

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