Integrate the following w.r.t.x.
`(2 sinx+3 cosx)/(3 sin x+4 cos x)`

Explanation

`"Let I"=int (2 sinx+3cos x)/(3 sin x +4 cos x)dx`
Put Numerator=A(Denominator)+B `=[(d)/(dx)("Denominator")]`
`therefore 2sinx+3cosx=A(3sinx+4cosx)+B[(d)/(dx)(3 sinx+4 cos)]`
`=A(3sin x+4cosx)+B(3cosx-4sinx)`
`=(3A-4B)sinx+(4A+3B)cos x`
Equating the coefficients of sin x and cos x on both sides, we get,
3A-4B=2 .......(1)
and 4A+3B=3 .......(2)
Multiplying equation (1) by 3 and equation (2) by 4, we get
`9A-12B=6`
`16A+12B=12`
On addding, we get
25A=18
`therefore A=(18)/(25)`
`therefore` from (1) `3((18)/(25))-4B=2`
`therefore 4B=(54)/(25)-2=(4)/(25)`
`therefore B=(1)/(25)`
`therefore 2sinx+3cosx=(18)/(25)(3 sin x+4cos x)+(1)/(25)(3cosx-4sinx`
`therefore I=int[((18)/(25)(3sinx+4cosx)+(1)/(25)(3cosx-4sinx))/(3sinx+4cosx)]dx`
`=int[(18)/(25)+(1)/(25)((3cosx-4sinx)/(3sinx+4cosx))]dx`
`=(18)/(25) int 1dx+(1)/(25) int(3cosx-4sinx)/(3sinx+4cos x)dx`
`=(18)/(25)x+(1)/(25)log|3sinx+4cos x|+c`
`.........[therefore int (f'(x))/(f(x))dx=log|f(x)|+c]`