plz evaluate the given integral solve each step and clear

let I= int 0_2pi 1/(a+bsinx)^2 dx

f(x) can be replaced by f(a+b-x)

so I= int 0_2pi 1/(a+bsin(2pi-x))^2 dx= int 0_2pi 1/(a – bsinx)^2 dx

adding we get 2I= int 0_2pi 2(a^2+b^2sin^2x)/(a^2 – b^2sin^2x)^2 dx

or I=  a^2*int 0_2pi 1/(a^2 – b^2sin^2x)^2 dx+ b^2*int 0_2pi sin^2x/(a^2 – b^2sin^2x)^2 dx

now both the above integrals are of the standard form and can be easily solved by multiplying the numerator and denominator by sec^4x and then substituting y= tanx. i know this ques is super complicated but thats the way it is :(

           an easier approach can be to write sinx in terms of tanx/2 and then substituting u= tanx/2

kindly approve :)