Let
3cosx+2sinx=A(4sinx+5cosx)+Bddx(4sinx+5cosx)=A(4sinx+5cosx)+B(4cosx−5sinx)
Comparing the coefficients of sinx and cosx, we get
4A−5B=25A+4B=3
Solving, we get
A=2341 and B=241
Thus the given integral reduces to
I=2341∫dx+241∫4cosx−5sinx4sinx+5cosxdx =2341x+241ln|4sinx+5cosx|+C
Thus,
a−2b=1941ab=461681