Question Number 102690 by bramlex last updated on 10/Jul/20

(1)∫(1/(cos (√x))) dx   (2) ∫ (1/(2+cot x)) dx   (3) ∫ (1/(ln(cos x))) dx

Answered by Dwaipayan Shikari last updated on 10/Jul/20

2)∫((tanx)/(2tanx+1))=(1/2)∫((2tanx+1)/(2tanx+1))−(1/2)∫(1/(2tanx+1))=(x/2)−(1/2)∫(2/((2t+1)(t^2 +1)))dt    =(x/2)−∫((At+C)/(t^2 +1))+(B/(2t+1))          {(1/((2t+1)(t^2 +1)))=((At+C)/(t^2 +1))+(B/(2t+1))  and take  (t=tanx)  =(x/2)−I_a                                     {2At^2 +At+2Ct+C+Bt^2 +B=1  =(x/2)+∫(((2/5)t−(1/5))/(t^2 +1))−((4/5)/(2t+1))             {  2A+B=0    A+2C=0  B+C=1  =(x/2)+(1/5)∫((2t−1)/(t^2 +1))−(4/(10))∫(1/(t+(1/2)))dt     {A=−(2/5)  B=(4/5)  C=(1/5)  =(x/2)+(1/5)∫((2t)/(t^2 +1))−(1/5)tan^(−1) tanx−(4/(10))log(t+(1/2))  =(x/2)+(1/5)log(t^2 +1)−(1/5)x−(4/(10))log(tanx+(1/2))  =(x/2)+(2/5)log(secx)−(1/5)x−(4/(10))log(tanx+(1/2))+Constant  =(1/5)(2x−log(cosx)−2log(tanx+(1/2)))+Constant

Answered by mathmax by abdo last updated on 10/Jul/20

2) I =∫  (dx/(2+((cosx)/(sinx)))) ⇒ I =∫  ((sinx)/(2sinx +cosx))dx changement tan((x/2))=t give  I =∫  (((2t)/(1+t^2 ))/(2×((2t)/(1+t^2 )) +((1−t^2 )/(1+t^2 )))) dt =∫  ((2t)/(4t+1−t^2 )) dt =−2 ∫  ((tdt)/(t^2 −4t−1))  t^2 −4t−1 =0 →Δ^′  =4+1 =5 ⇒t_1 =2+(√5) and t_2 =2−(√5)  ⇒ F(t) =(t/(t^2 −4t−1)) =(t/((t−t_1 )(t−t_2 ))) =(a/(t−t_1 )) +(b/(t−t_2 ))  a =(t_1 /(t_1 −t_2 )) =((2+(√5))/(2(√5))) , b =(t_2 /(t_2 −t_1 )) =((2−(√5))/(−2(√5))) =((−2+(√5))/(2(√5))) ⇒  I =−2 ∫   ((adt)/(t−t_1 )) −2 ∫ ((bdt)/(t−t_2 ))  =−2aln∣t−t_1 ∣ −2 b ln∣t−t_2 ∣ +c  =−2×((2+(√5))/(2%5))ln∣tan((x/2))−2−(√5)∣−2×((−2+(√5))/(2(√5)))ln∣tan((x/2))−2+(√5)∣ +C  =(((−2−(√5))/(√5)))ln∣tan((x/2))−2−(√5)∣ +((2−(√5))/(√5))ln∣tan((x/2))−2+(√5)∣ +C

Answered by bemath last updated on 11/Jul/20