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Question Number 46898 by  last updated on 02/Nov/18

∫((tanx)/((tanx+1)^2 −2tan^2 x ))dx=??

∫tanx(tanx+1)2−2tan2xdx=??

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 .

changementtanx=tgiveI=∫t((t+1)2−2t2)dt(1+t2)=∫tdt(t2+2t+1−2t2)(1+t2)=∫tdt(−t2+2t+1)(t2+1)=−∫t(t2+1)(t2−2t−1)letdecomposeF(t)=t(t2+1)(t2−2t−1)t2−2t−1=0⇒t2−2t+1−2=0⇒(t−1)2−2=0⇒t1=1+2andt2=1−2F(t)=t(t−t1)(t−t2)(t2+1)=at−t1+bt−t2+ct+dt2+1a=limt→t1(t−t1)F(t)=t1(t1−t2)(1+t12)=1+222(1+3+22)=1+222(4+22)=1+282+8=18b=limt→t2(t−t2)F(t)=t2(t2−t1)(1+t22)=1−2−22(1+3−22)=1−2−22(4−22)=−1−282−8=2−18(2−1)=18⇒F(t)=18(t−t1)+18(t−t2)+bt+ct2+1F(0)=0=−18t1−18t2+c⇒c=18(t1+t2t1t2)=18(2−1)=−14limt→+∞tF(t)=0=14+b⇒b=−14⇒F(t)=18(t−t1)+18(t−t2)−14t+1t2+1⇒∫F(t)dt=18ln∣t−t1∣+18ln∣t−t2∣−18ln(t2+1)−14arctan(t)+c⇒I=−18ln∣t−(1+2))∣−18ln∣1−(1−2)∣+18ln(t2+1)+14arctant+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 .

⇒I=−18ln∣(t−t1)(t−t2)∣+18ln(t2+1)+14arctant+c=−18ln∣t2−2t−1∣+18ln(t2+1)+arctant4+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 .

lettanx=tI=∫tdt(1+t2)(1+2t−t2)lett(1+t2)(1+2t−t2)=At+B1+t2+Bt+C1+2t−t2⇒t=(At+B)(1+2t−t2)+(Bt+C)(1+t2)⇒−A+B=0−B+2A+C=02B+A+B=1B+C=0⇒A=B=−C=14I=14∫t+1t2+1dt+∫t−12−(t−1)2dtI=18ln∣t2+1∣+14tan−1t−18ln∣1+2t−t2∣+ctbeingtanx.

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

12∫2tanxtan2x+1+2anx−2tan2xdx12∫2tanx2tanx+1−tan2xdxdivideby1+tan2x12∫2tanx1+tan2x2tanx1+tan2x+1−tan2x1+tan2xdx12∫sin2xsin2x+cos2xdx14∫sin2x+sin2x−cos2x+cos2xsin2x+cos2xdx14∫sin2x+cos2xsin2x+cos2xdx−14∫cos2x−sin2xsin2x+cos2xdx14x−14ln∣sin2x+cos2x∣+C

Commented by  last updated on 04/Nov/18

Very Good Answer Sir It Is Correct ! Thanku Sir

VeryGoodAnswerSirItIsCorrect!ThankuSir

Commented by  last updated on 04/Nov/18

Sorry... (1/4)x−(1/2)ln∣sin2x+cos2x∣+C

Sorry...14x−12ln∣sin2x+cos2x∣+C

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