The value of `int(1)/(3 sin x -cos x +3)dx` is equal to

Options

A. `log((tan.(x)/(2)+1)/(2tan.(x)/(2)+1))+C`
B. `(1)/(2)log((2tan.(x)/(2)+1)/(tan.(x)/(2)+1))+C`
C. `log((2log.(x)/(2)+1)/(tan.(x)/(2)+1))+C`
D. `2log((2tan.(x)/(2)+1)/(tan.(x)/(2)+1))+C`

Answer

C

Explanation

Let `" "l=int(dx)/(3sin x- cos x+3)`
`{because sinx=(2tan.(x)/(2))/(1+tan^(2).(x)/(2))and cosx=(1-tan^(2).(x)/(2))/(1+tan^(2).(x)/(2))}`
`l=int(dx)/(3{(2tan.(x)/(2))/(1+tan^(2).(x)/(2))}-{(1-tan^(2).(x)/(2))/(1+tan^(2).(x)/(2))}+3)`
`=int((1+tan^(2).(x)/(2))dx)/(6 tan.(x)/(2)-1+tan^(2).(x)/(2)+3+3 tan^(2).(x)/(2))`
`=int(sec^(2).(x)/(2))/(4tan^(2).(x)/(2)+6tan.(x)/(2)+2)dx`
`" "["let "t=tan.(x)/(2),dt=(1)/(2)sec^(2).(x)/(2)dx]`
`=int(dt)/(2t^(2)+3t+1)=int(dt)/((t+1)(2t+1))`
`=int{(-1)/((t+1))+(2)/((2t+1))}dt" [by partial fraction]"`
`=-log(t+1)+(2)/(2)log(2t+1)+C`
`=(log(2t+1))/((t+1))+C=log((2tan.(x)/(2)+1)/(tan.(x)/(2)+1))+C`