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Question Number 71814 by Aman Arya last updated on 20/Oct/19

∫(1/(1+cot x))dx

11+cotxdx

Commented by mathmax by abdo last updated on 20/Oct/19

let I=∫ (dx/(1+cotanx)) ⇒I=∫ (dx/(1+((cosx)/(sinx)))) =∫((sinx)/(sinx +cosx))dx  changement tan((x/2))=t give I=∫  (((2t)/(1+t^2 ))/(((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 ))))((2dt)/(1+t^2 ))  I=∫    ((2t)/((1+t^2 )^2 {((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 ))}))dt =∫ ((2t)/((1+t^2 )(−t^2 +2t+1)))dt  =−2∫(t/((t^2 +1)(t^2 −2t−1)))dt let decompose F(t)=(t/((t^2 +1)(t^2 −2t−1)))  t^2 −2t−1=0→Δ^′ =1+1=2 ⇒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 )(t_1 ^2  +1))) =((1+(√2))/(2(√2)(3+2(√2)+1)))  =((2+(√2))/(4(4+2(√2)))) =((2+(√2))/(8(2+(√2)))) =(1/8)  b=(t_2 /((t_2 −t_1 )(t_2 ^2 +1))) =((1−(√2))/((−2(√2))(3−2(√2)+1))) =((−1+(√2))/(2(√2)(4−2(√2))))  =((−(√2)+2)/(8(2−(√2)))) =(1/8)  lim_(t→+∞) tF(t)=0=a+b+c ⇒c=−a−b =−(1/4) ⇒  F(t)=(1/(8(t−t_1 ))) +(1/(8(t−t_2 ))) +((−(1/4)t+d)/(t^2  +1))  F(0)=0 =−(1/(8t_1 ))−(1/(8t_2 )) +d ⇒d=(1/8)((1/t_1 )+(1/t_2 ))=(1/8)(((t_1 +t_2 )/(t_1 t_2 )))  =(1/8)(2/(−1)) =−(1/4) ⇒F(t)=(1/(8(t−t_1 ))) +(1/(8(t−t_2 )))−(1/4) ((t+1)/(t^2  +1)) ⇒  I =−(1/4)∫  (dt/(t−t_1 ))−(1/4)∫ (dt/(t−t_2 )) +(1/2)∫  ((t+1)/(t^2  +1))dt  =−(1/4)ln∣t−t_1 ∣−(1/4)ln∣t−t_2 ∣+(1/4)ln(t^2 +1) +(1/2)arctant +c  =−(1/4)ln∣tan((x/2))−1−(√2)∣−(1/4)ln∣tan((x/2))−1+(√2)∣  +(1/4)ln(1+tan^2 ((x/2))) +(x/4) +c .

letI=dx1+cotanxI=dx1+cosxsinx=sinxsinx+cosxdxchangementtan(x2)=tgiveI=2t1+t22t1+t2+1t21+t22dt1+t2I=2t(1+t2)2{2t1+t2+1t21+t2}dt=2t(1+t2)(t2+2t+1)dt=2t(t2+1)(t22t1)dtletdecomposeF(t)=t(t2+1)(t22t1)t22t1=0Δ=1+1=2t1=1+2andt2=12F(t)=t(tt1)(tt2)(t2+1)=att1+btt2+ct+dt2+1a=limtt1(tt1)F(t)=t1(t1t2)(t12+1)=1+222(3+22+1)=2+24(4+22)=2+28(2+2)=18b=t2(t2t1)(t22+1)=12(22)(322+1)=1+222(422)=2+28(22)=18limt+tF(t)=0=a+b+cc=ab=14F(t)=18(tt1)+18(tt2)+14t+dt2+1F(0)=0=18t118t2+dd=18(1t1+1t2)=18(t1+t2t1t2)=1821=14F(t)=18(tt1)+18(tt2)14t+1t2+1I=14dttt114dttt2+12t+1t2+1dt=14lntt114lntt2+14ln(t2+1)+12arctant+c=14lntan(x2)1214lntan(x2)1+2+14ln(1+tan2(x2))+x4+c.

Commented by petrochengula last updated on 21/Oct/19

=∫(1/(1+((cosx)/(sinx))))dx  =∫((sinx)/(sinx+cosx))dx  =(1/2)∫((2sinx)/(sinx+cosx))dx  =(1/2)∫((sinx+cosx)/(sinx+cosx))dx−(1/2)∫((cosx−sinx)/(sinx+cosx))dx  =(1/2)x−(1/2)ln∣sinx+cosx∣+C

=11+cosxsinxdx=sinxsinx+cosxdx=122sinxsinx+cosxdx=12sinx+cosxsinx+cosxdx12cosxsinxsinx+cosxdx=12x12lnsinx+cosx+C

Answered by $@ty@m123 last updated on 20/Oct/19

∫(1/(1+((cos  x)/(sin  x))))dx  ∫((sin x)/(sin x+cos x))dx  ∫(((1/2)(sin x+cos x)−(1/2)(cos x−sin  x))/(sin x+cos x))dx  =(1/2)∫dx−(1/2)∫((d(sin x+cos x))/(sin x+cos x))  =(1/2)x−(1/2)ln (sin x+cos x)+C

11+cosxsinxdxsinxsinx+cosxdx12(sinx+cosx)12(cosxsinx)sinx+cosxdx=12dx12d(sinx+cosx)sinx+cosx=12x12ln(sinx+cosx)+C

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