Question Number 102198 by Study last updated on 07/Jul/20

∫(x/(sin^2 x−3))dx=?

Answered by mathmax by abdo last updated on 08/Jul/20

not resoluble(√!)

Answered by Dwaipayan Shikari last updated on 08/Jul/20

∫(x/((sinx+(√3))(sinx−(√3))))  (x/(2(√3)))∫(1/(sinx−(√3)))−(1/(sinx−(√3)))dx−∫(∫(1/(sin^2 x−3)))dx  x(1/(2(√3)))(I_a −I_b )−∫(∫(1/(sin^2 x−3)))dx  I_a =∫(1/(sinx−(√3)))dx      =2∫(dt/((((2t)/(t^2 +1))−(√3))(t^2 +1)))=2∫(1/(2t−(√3)t^2 −(√3)))dt=−(2/(√3))∫(1/(t^2 −((2t)/(√3))+1))                                                                                               =−(2/(√3))∫(1/((t−(1/(√3)))^2 +((√(2/3)))^2 ))dt                                                                                =−(1/(√2))tan^(−1) (((√3)t−1)/(√2))                                                                                =−(1/(√2))tan^(−1) (((√3)tan(x/2)−1)/(√2))  I_b =∫(1/(sinx+(√3)))=2∫(1/((((2t)/(t^2 +1))+(√3))(t^2 +1)))dt  =(2/(√3))∫(1/(t^2 +((2t)/(√3))+1))=(1/(√2))tan^(−1) (((√3)tan(x/2)+1)/(√2))  (x/(2(√3)))(I_a −I_b )=−(x/(2(√6)))(tan^(−1) (((√3)tan(x/2)+1)/(√2))+tan^(−1) (((√3)tan(x/2)−1)/(√2)))....  .....continue        But now we have to integrate  ∫(I_a −I_b )= −∫tan^(−1) ((((2(√3)tan(x/2))/(√2))/(1−((3tan^2 (x/2)−1)/2))))  −∫tan^(−1) (((2(√6)tan(x/2))/(1−3tan^2 (x/2))))........continue  =−xtan^(−1) (((2(√6)tan(x/2))/(1−3tan^2 (x/2))))+∫((x(√6)sec^2 (x/2))/(1+9tan^4 (x/2)+18tan^2 (x/2)))      (x/(−2(√3)))(I_a −I_b )−∫(I_a −I_b )  =−(((x+1))/(2(√6)))(tan^(−1) (((2(√6)tan(x/2))/(1−3tan^2 (x/2)))))+∫((x(√6)sec^2 (x/2))/(1+9tan^4 (x/2)+18tan^2 (x/2)))+Constant