Let I = ∫ csc x/(3cos x+4sin x) dx. Divide numerator and denominator by sin x: I = ∫ csc²x/(3cot x+4) dx. Put t = 3cot x+4, so dt = -3csc²x dx. Then I = -1/3 ∫ dt/t = -1/3 log|t| + C = 1/3 log|1/t| + C = 1/3 log|sin x/(3cos x+4sin x)| + C. Hence option 1 is correct.