Question Number 103825 by mohammad17 last updated on 17/Jul/20

∫ (dx/((1−sinx)^2 )) ?

Answered by ~blr237~ last updated on 17/Jul/20

 ∫ ((1+2sinx+(1−cos^2 x))/(cos^4 x))dx  ∫ [ 2(1+tan^2 x)+2(1+tan^2 x)tan^2 x +((2sinx)/((cosx)^4 )) −(1/(cos^2 x)))dx

Commented bymohammad17 last updated on 17/Jul/20

how this sir

Answered by mathmax by abdo last updated on 17/Jul/20

let f(a) =∫  (dx/(a−sinx))  we have f^′ (a) =−∫  (dx/((a−sinx)^2 )) ⇒  ∫ (dx/((a−sinx)^2 )) =−f^′ (a) and ∫ (dx/((1−sinx)^2 )) =−f^′ (1) let explicite f(a)  changement tan((x/2))=t give f(a) =∫  ((2dt)/((1+t^2 )(a−((2t)/(1+t^2 )))))  =∫ ((2dt)/(a+at^2 −2t)) =∫ ((2dt)/(at^2 −2t +a))  Δ^′  =1−a^2  let take a>1 ⇒ Δ^′ <0 ⇒f(a) =(2/a)∫ (dt/(t^2 −((2t)/a) +1))  =(2/a)∫  (dt/(t^2 −2(t/a) +(1/a^2 ) +1−(1/a^2 ))) =(2/a)∫  (dt/((t−(1/a))^2  +((a^2 −1)/a^2 )))  =_(t−(1/a)=((√(a^2 −1))/a)u)     (2/a)×(a^2 /(a^2 −1)) ∫  (1/(1+u^2 ))×((√(a^2 −1))/a)du  =(2/(√(a^2 −1)))∫ (du/(1+u^2 )) =(2/(√(a^2 −1))) arctan(((at−1)/(√(a^2 −1)))) =(2/(√(a^2 −1))) arctan(((atan((x/2))−1)/(√(a^2 −1)))) +c  rest to calculate f^′ (a) and lim_(a→1) f^′ (a) ...be continued...

Answered by Dwaipayan Shikari last updated on 17/Jul/20

∫(((1+sinx)^2 )/((1−sin^2 x)^2 ))dx=∫(sec^2 x+secxtanx)^2 dx=∫sec^2 x(secx+tanx)^2 dx  ∫secx(secx+tanx)secx(secx+tanx)dx  ∫secx  tdt                             {        take  secx+tanx=t  (dt/dx)=secx(secx+tanx)   ∫(t/(√(2t^2 −t^4 )))   dt                       {   ((1+sinx)/(cosx))=t⇒1+sin^2 x+2sinx=t^2 (1−sin^2 x)  ∫(1/(√(2−t^2 )))dt                              {        ⇒(1+t^2 )sin^2 x+2sinx+1−t^2 =0  ∫(((√2) cosθ)/(√(2−2sin^2 θ)))dθ                {t=(√2)sinθ    (√2)cosθ=(dt/dθ)  ∫dθ                                                 {       sinx=((−2±(√(4−4(1−t^4 ))))/2)=t^2 −1  =θ+C                                                         {      cosx=(√(1−(t^2 −1)^2 ))  =sin^(−1) (t/(√2))+C  =sin^(−1) (((secx+tanx)/(√2)))+Constant

Commented bymohammad17 last updated on 17/Jul/20

thank you sir

Commented byDwaipayan Shikari last updated on 17/Jul/20

Kindly check it

Commented byDwaipayan Shikari last updated on 17/Jul/20

Commented byDwaipayan Shikari last updated on 17/Jul/20

Alternate solution

Answered by mathmax by abdo last updated on 17/Jul/20

let try another way   I =∫  (dx/((1−sinx)^2 )) ⇒ I =∫ (dx/((1−cos((π/2)−x))^2 )) =∫ (dx/((2sin^2 ((π/4)−(x/2)))^4 ))  =(1/(16))∫  (dx/(sin^4 ((π/4)−(x/2)))) =_((π/4)−(x/2)=t)    (1/(16)) ∫  ((−2dt)/(sin^4 t)) =−(1/8)∫ (dt/(sin^4 t))  changement  tan((t/2))=u give ∫ (dt/(sin^4 t)) =∫  ((2du)/((1+u^2 )(((2u)/(1+u^2 )))^4 ))  =∫ ((2(1+u)^3 )/(16u^4 )) du =(1/8) ∫ ((u^3  +3u^2  +3u +1)/u^4 )du  =(1/8)∫ ((1/u) +(3/u^2 ) +(3/u^3 ) +(1/u^4 ))du  =(1/8){ ln∣u∣−(3/u) +3(−(1/(2u^2 ))) −(1/(3u^3 ))} +C  ⇒ I =−(1/(64)){ ln∣tan((t/2))∣−(3/(tan((t/2)))) −(3/(2tan^2 ((t/2))))−(1/(3tan^3 ((t/2))))} +C  =−(1/(64)){ ln∣tan((π/8)−(x/4))∣−(3/(tan((π/8)−(x/4)))) −(3/(2tan^2 ((π/8)−(x/4))))−(1/(3tan^3 ((π/8)−(x/4))))} +C

Commented bymathmax by abdo last updated on 17/Jul/20

error of typo I =∫ (dx/((2sin^2 ((π/4)−(x/2)))^2 )) =...

Commented bymathmax by abdo last updated on 17/Jul/20

sorry change 64 by 32...

Commented bymohammad17 last updated on 17/Jul/20

ok sir thank you

Answered by PRITHWISH SEN 2 last updated on 17/Jul/20

∫(dx/((1−((2tan (x/2))/(1+tan ^2 (x/2))))^2 ))=∫((sec^2 (x/2))/((1−tan (x/2))^4 ))(1+tan ^2 (x/2))dx  tan(x/2)=t⇒sec ^2 (x/2)dx=2dt  =2∫(((1+t^2 )dt)/((1−t)^4 ))=∫((2dt)/((1−t)^2 ))−∫((4(1−t)dt)/((1−t)^4 ))+4∫(dt/((1−t)^4 ))  =(2/((1−t)))+(2/((1−t)^2 ))−(4/(3(1−t)^3 )) +C  = (2/((1−tan (x/2)))) +(2/((1−tan (x/2))^2 )) −(4/(3(1−tan (x/2))^3 )) +C  please check

Commented bymohammad17 last updated on 17/Jul/20

than you sir

Answered by Ar Brandon last updated on 19/Jul/20

I=∫(dx/((1−sinx)^2 ))=∫(((1+sinx)^2 )/((1−sinx)^2 (1+sinx)^2 ))dx     =∫((1+2sinx+sin^2 x)/((1−sin^2 x)^2 ))dx=∫((cos^2 x+2sinx+2sin^2 x)/(cos^4 x))dx     =∫sec^2 xdx+2∫((sinx)/(cos^4 x))dx+2∫tan^2 xsec^2 xdx     =tanx+((2sec^3 x)/3)+((2tan^3 x)/3)+C

Answered by Her_Majesty last updated on 07/Aug/20

∫(dx/((1−sin x)^2 ))  simply Weyerstrass′ Substitution  t=tan (x/2)  leads to  2∫((t^2 +1)/((t−1)^4 ))dt=−((2(3t^2 −3t+2))/(3(t−1)^3 ))=...