{(2cosx+3)/(2+3cosx)^2} dx

Solution: Let I = ∫ 3 + 2 cos x / (2+ 3 cos x)² dx 

Multiplying Nr. & Dr. by cosec²x

=> I = int {{(3cosec^{2}x+ 2cot x cosec x) } / (2cosec x +3cot x)^2  }dx


        = - int {{(-3cosec^{2}x- 2cot x cosec x) } / (2cosec x +3cot x)^2  }dx

  Let us consider 2cosec x +3cot x=t

  then    dt = (-3cosec^{2}x- 2cot x cosec x).dx

         =int (dt / t^2)

        integrate it then we get

        ={1} / {(2cosec x +3cot x)}


        ={sin x}  / {2+3cos x} +c

given ∫ {(2cosx+3)/(2+3cosx)^2} dx

I =∫{(2cosx +3)/(2+3cosx)^2} dx

let multiplying N.R and D.R  with cosec^2 x

then we get

I = - ∫ {(-2cosec x.cotx - 3cosec^2 x)/(2cosecx+3cotx)^2}.dx

let us consider t=2cosecx+3cotx

then dt =( -2cosec x.cotx - 3cosec^2 x ).dx

      = -∫-(1/t^2) => 1/t  => 1/(2cosecx+3cotx.cotx)

      ={(sinx) /(2+3cosx)}