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Question Number 46898 by  last updated on 02/Nov/18
∫((tanx)/((tanx+1)^2 −2tan^2 x ))dx=??
$$\int\frac{{tanx}}{\left({tanx}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{tan}^{\mathrm{2}} {x}\:\:}{dx}=?? \\ $$
Commented by prof Abdo imad last updated on 02/Nov/18
changement tanx =t give I = ∫ (t/(((t+1)^2 −2t^2 ))) (dt/((1+t^2 ))) =∫ ((tdt)/((t^2 +2t+1−2t^2 )(1+t^2 ))) = ∫ ((tdt)/((−t^2 +2t+1)(t^2 +1))) =−∫ (t/((t^2 +1)(t^2 −2t−1))) let decompose F(t)=(t/((t^2 +1)(t^2 −2t−1))) t^2 −2t−1 =0 ⇒t^2 −2t +1−2 =0 ⇒ (t−1)^2 −2 =0 ⇒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 )) +((ct +d)/(t^2 +1)) a =lim_(t→t_1 ) (t−t_1 )F(t) = (t_1 /((t_1 −t_2 )(1+t_1 ^2 ))) =((1+(√2))/(2(√2)(1 +3 +2(√2)))) =((1+(√2))/(2(√2)(4+2(√2)))) =((1+(√2))/(8(√2)+8)) =(1/8) b =lim_(t→t_2 ) (t−t_2 )F(t) =(t_2 /((t_2 −t_1 )(1+t_2 ^2 ))) =((1−(√2))/(−2(√2)(1+3−2(√2)))) =((1−(√2))/(−2(√2)(4−2(√2)))) =−((1−(√2))/(8(√2)−8)) =(((√2)−1)/(8((√2)−1))) =(1/8) ⇒ F(t)= (1/(8(t−t_1 ))) +(1/(8(t−t_2 ))) +((bt +c)/(t^(2 ) +1)) F(0) =0 =((−1)/(8t_1 )) −(1/(8t_2 )) +c ⇒c =(1/8)(((t_1 +t_2 )/(t_1 t_2 ))) =(1/8)((2/(−1)))=−(1/4) lim_(t→+∞) t F(t) =0 =(1/4) +b ⇒b=−(1/4) ⇒ F(t) =(1/(8(t−t_1 ))) +(1/(8(t−t_2 ))) −(1/4) ((t+1)/(t^2 +1)) ⇒ ∫ F(t)dt =(1/8)ln∣t−t_1 ∣+(1/8)ln∣t−t_2 ∣−(1/8)ln(t^2 +1) −(1/4)arctan(t)+c ⇒ I =−(1/8)ln∣t−(1+(√2)))∣−(1/8)ln∣1−(1−(√2))∣ +(1/8)ln(t^2 +1)+(1/4) arctant +C .
$${changement}\:{tanx}\:={t}\:{give} \\ $$$${I}\:=\:\int\:\:\frac{{t}}{\left(\left({t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} \right)}\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\int\:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$=\:\int\:\:\:\frac{{tdt}}{\left(−{t}^{\mathrm{2}} \:+\mathrm{2}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\:=−\int\:\frac{{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\frac{{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\:=\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} −\mathrm{2}{t}\:+\mathrm{1}−\mathrm{2}\:=\mathrm{0}\:\Rightarrow \\ $$$$\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}\:=\mathrm{0}\:\Rightarrow{t}_{\mathrm{1}} \:=\mathrm{1}+\sqrt{\mathrm{2}}\:\:{and}\:{t}_{\mathrm{2}} =\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$${F}\left({t}\right)=\frac{{t}}{\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\:=\frac{{a}}{{t}−{t}_{\mathrm{1}} }\:+\frac{{b}}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{t}\rightarrow{t}_{\mathrm{1}} } \left({t}−{t}_{\mathrm{1}} \right){F}\left({t}\right)\:=\:\frac{{t}_{\mathrm{1}} }{\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)\left(\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}\:+\mathrm{3}\:+\mathrm{2}\sqrt{\mathrm{2}}\right)}\:=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\right)}\:=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{8}\sqrt{\mathrm{2}}+\mathrm{8}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${b}\:={lim}_{{t}\rightarrow{t}_{\mathrm{2}} } \left({t}−{t}_{\mathrm{2}} \right){F}\left({t}\right)\:=\frac{{t}_{\mathrm{2}} }{\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left(\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{−\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}\:=\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{−\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)} \\ $$$$=−\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{8}}\:=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{8}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$${F}\left({t}\right)=\:\frac{\mathrm{1}}{\mathrm{8}\left({t}−{t}_{\mathrm{1}} \right)}\:+\frac{\mathrm{1}}{\mathrm{8}\left({t}−{t}_{\mathrm{2}} \right)}\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}\:} +\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\frac{−\mathrm{1}}{\mathrm{8}{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{\mathrm{8}{t}_{\mathrm{2}} }\:+{c}\:\Rightarrow{c}\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{{t}_{\mathrm{1}} +{t}_{\mathrm{2}} }{{t}_{\mathrm{1}} {t}_{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{2}}{−\mathrm{1}}\right)=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${lim}_{{t}\rightarrow+\infty} {t}\:{F}\left({t}\right)\:=\mathrm{0}\:=\frac{\mathrm{1}}{\mathrm{4}}\:+{b}\:\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{8}\left({t}−{t}_{\mathrm{1}} \right)}\:+\frac{\mathrm{1}}{\mathrm{8}\left({t}−{t}_{\mathrm{2}} \right)}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int\:{F}\left({t}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid{t}−{t}_{\mathrm{1}} \mid+\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid{t}−{t}_{\mathrm{2}} \mid−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}{arctan}\left({t}\right)+{c}\:\Rightarrow \\ $$$$\left.{I}\:\:=−\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid{t}−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right)\mid−\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid\mathrm{1}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mid \\ $$$$+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{4}}\:{arctant}\:+{C}\:. \\ $$
Commented by prof Abdo imad last updated on 02/Nov/18
⇒ I =−(1/8)ln∣(t−t_1 )(t−t_2 )∣+(1/8)ln(t^2 +1)+(1/4)arctant +c =−(1/8)ln∣t^2 −2t−1∣+(1/8)ln(t^2 +1)+((arctant)/4) +C .
$$\Rightarrow\:{I}\:=−\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)\mid+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{4}}{arctant}\:+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)+\frac{{arctant}}{\mathrm{4}}\:+{C}\:. \\ $$
Answered by ajfour last updated on 02/Nov/18
let tan x=t I=∫((tdt)/((1+t^2 )(1+2t−t^2 ))) let (t/((1+t^2 )(1+2t−t^2 )))=((At+B)/(1+t^2 )) +((Bt+C)/(1+2t−t^2 )) ⇒ t=(At+B)(1+2t−t^2 ) +(Bt+C)(1+t^2 ) ⇒ −A+B=0 −B+2A+C = 0 2B+A+B=1 B+C=0 ⇒ A=B=−C = (1/4) I=(1/4)∫((t+1)/(t^2 +1))dt+∫((t−1)/(2−(t−1)^2 ))dt I = (1/8)ln ∣t^2 +1∣+(1/4)tan^(−1) t −(1/8)ln ∣1+2t−t^2 ∣+c t being tan x .
$${let}\:\mathrm{tan}\:{x}={t} \\ $$$${I}=\int\frac{{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right)} \\ $$$${let}\:\frac{{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right)}=\frac{{At}+{B}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{Bt}+{C}}{\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{t}=\left({At}+{B}\right)\left(\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({Bt}+{C}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:−{A}+{B}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:−{B}+\mathrm{2}{A}+{C}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{B}+{A}+{B}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{B}+{C}=\mathrm{0} \\ $$$$\Rightarrow\:{A}={B}=−{C}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}+\int\frac{{t}−\mathrm{1}}{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$\:\:{I}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid{t}^{\mathrm{2}} +\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}^{−\mathrm{1}} {t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid\mathrm{1}+\mathrm{2}{t}−{t}^{\mathrm{2}} \mid+{c} \\ $$$$\:\:\:\:\:\:{t}\:{being}\:\mathrm{tan}\:{x}\:. \\ $$
Answered by  last updated on 04/Nov/18
(1/2)∫((2tanx)/(tan^2 x+1+2anx−2tan^2 x))dx (1/2)∫((2tanx)/(2tanx+1−tan^2 x))dx divide by 1+tan^2 x (1/2)∫(((2tanx)/(1+tan^2 x))/(((2tanx)/(1+tan^2 x))+((1−tan^2 x)/(1+tan^2 x))))dx (1/2)∫((sin2x)/(sin2x+cos2x))dx (1/4)∫((sin2x+sin2x−cos2x+cos2x)/(sin2x+cos2x))dx (1/4)∫((sin2x+cos2x)/(sin2x+cos2x))dx−(1/4)∫((cos2x−sin2x)/(sin2x+cos2x))dx (1/4)x−(1/(4 ))ln∣sin2x+cos2x∣+C
$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{tanx}}{{tan}^{\mathrm{2}} {x}+\mathrm{1}+\mathrm{2}{anx}−\mathrm{2}{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{tanx}}{\mathrm{2}{tanx}+\mathrm{1}−{tan}^{\mathrm{2}} {x}}{dx} \\ $$$${divide}\:{by}\:\mathrm{1}+{tan}^{\mathrm{2}} {x} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}{\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}+\frac{\mathrm{1}−{tan}^{\mathrm{2}} {x}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{sin}\mathrm{2}{x}}{{sin}\mathrm{2}{x}+{cos}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{sin}\mathrm{2}{x}+{sin}\mathrm{2}{x}−{cos}\mathrm{2}{x}+{cos}\mathrm{2}{x}}{{sin}\mathrm{2}{x}+{cos}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{sin}\mathrm{2}{x}+{cos}\mathrm{2}{x}}{{sin}\mathrm{2}{x}+{cos}\mathrm{2}{x}}{dx}−\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{cos}\mathrm{2}{x}−{sin}\mathrm{2}{x}}{{sin}\mathrm{2}{x}+{cos}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{x}−\frac{\mathrm{1}}{\mathrm{4}\:}{ln}\mid{sin}\mathrm{2}{x}+{cos}\mathrm{2}{x}\mid+{C} \\ $$$$ \\ $$
Commented by  last updated on 04/Nov/18
Very Good Answer Sir It Is Correct ! Thanku Sir
$${Very}\:{Good}\:{Answer}\:{Sir}\: \\ $$$${It}\:{Is}\:{Correct}\:! \\ $$$${Thanku}\:\:{Sir} \\ $$
Commented by  last updated on 04/Nov/18
Sorry... (1/4)x−(1/2)ln∣sin2x+cos2x∣+C
$${Sorry}… \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}{x}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{sin}\mathrm{2}{x}+{cos}\mathrm{2}{x}\mid+{C} \\ $$

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