Concept:Integration using substitution and partial fractions leads to comparing coefficients.Explanation:Let I=∫(1+sinx)(2+sinx)sin2xdx.Substitute t=sinx, so dt=cosxdx and sin2x=2sinxcosx=2tcosx.Then I=∫(t+1)(t+2)2tcosx⋅cosxdt=∫(t+1)(t+2)2tdt.Use partial fractions: (t+1)(t+2)2t=t+1A+t+2B.Multiply: 2t=A(t+2)+B(t+1)=(A+B)t+(2A+B).Equating coefficients: A+B=2 and 2A+B=0.Solve: subtract first from second → (2A+B)−(A+B)=0−2 → A=−2, then B=4.Thus I=∫(−t+12+t+24)dt=−2ln∣t+1∣+4ln∣t+2∣+C.Replace t=sinx: I=−2ln∣1+sinx∣+4ln∣2+sinx∣+C.The given form is I=aln∣1+sinx∣−bln∣2+sinx∣+C.Comparing: a=−2, and −b=4 → b=−4.Answer:a=−2, b=−4 (Option C).