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

∫(1/(1+cot x))dx

Commented bymathmax 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 .

Commented bypetrochengula 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

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