Evaluate the following integrals :
`(i) int(cosx)/(2sinx+3cosx)dx`
`(ii) int(2sinx+cosx)/(7sinx-5cosx)dx` ,br. `(iii) int(5sinx+6)/(2cosx+sinx+3)dx`

Explanation

`(i)` We have
`cosx=k(2sinx+3cosx)+m(2cosx-3sinx)`
`impliescosx=(2k-3m)sinx+(3k+2m)cosx`
Equating the coefficients `sinx` and `cosx` from both sides
`3k+2m=1`
`2k-3m=0`
when `k=(3)/(13)` and `m=(2)/(13)`
Thus `int(cosx)/(2sinx+3cosx)dx=(3)/(13)intdx+(2)/(13)int((2cosx-3sinx))/(2sinx+3cosx)dx`
`=(3)/(13)x+(2)/(13)ln|2sinx+3cosx|`
`(ii)` Let `2sinx+cosx=k(7sinx-5cosx)+m(7cosx+5sinx)`
`implies2sinx+cosx=(7k+5m)sinx+(7m-5k)cosx`
`implies7k+5m=2`
and `7m-5k=1` when `m=(17)/(74)` and `k=(9)/(14)`
Thus `int(2sinx+cosx)/(7sinx-5cosx)dx=(9)/(74)intdx+(17)/(74)int(7cosx+5sinx)/(7sinx-5cosx)dx`
`=(9)/(14)x+(17)/(74)ln|7sinx-5cosx|+C`
`(iii)` Let `5sinx+6=k(2cosx+sinx+3)+m(-2sinx+cosx)+n`
`implies5sinx+6=(2k+m)cosx+(k-2m)sinx+3k+n`
`implies2k+m=0`, `k-2m=5`, `3k+n=6`
`impliesk=1`, `n=3`, `m=-2`
Thus `int(5sinx+6)/(2cosx+sinx+3)dx=intdx-2int((cosx-2sinx))/(2cosx+sinx+3)dx+3int(1)/(2cosx+sinx+3)dx`
`=x-2ln|2cosx+sinx+3|+3l_(1)`, say
`l_(1)=int(1)/((2cosx+sinx+3)dx=int(1)/(2((1-tan^(2).(x)/(2))/(1+tan^(2).(x)/(2)))+(2tan.(x)/(2))/(1+tan^(2).(x)/(2))+3)dx`
`=int(1+tan^(2).(x)/(2))/(2-2tan^(2).(x)/(2)+2tan.(x)/(2)+3+3tan^(2).(x)/(2))dx`
`=int(sec^(2).(x)/(2))/(5+2tan.(x)/(2)+tan^(2).(x)/(2))dx`
Let us put `tan.(x)/(2)=u`
so that `sec^(2).(x)/(2)dx=2du`
`=int(2)/(4+(1+u)^(2))du`
`=2*(1)/(2)tan^(-1)((1+u)/(2))+C`
`=tan^(-1)((tan.(x)/(2)+1)/(2))+C`
Hence the given integral is
`int(5sinx+6)/(2cosx+sinx+3)dx=x-2ln|2cosx+sinx+3|+3tan^(-1)((sin.(x)/(2)+cos.(x)/(2))/(2cos.(x)/(2)))+C`