How to integrate the following problem dx/(3sinx+sin³x)

sin3x=3sinx-4sin3x => Sinx/Sin 3x = 1/(3-4 Sin2x)

so for integration of 1/(3-4 Sin2x), multiply & divide numerator & denominator by sec2x, we get

sec2x dx /(3sec2x -4 tan2x) =sec2x dx /(3(1+tan2x )-4 tan2x) = sec2x dx /(3 - tan2x) , now put tan2x = 't' which gives sec2x dx = dt, so we get integration of dt /(3-t2) = 1/2 /3  * log (/3 +t)//3+t)  + C .

Now put t = tanx in above result