Question

How do I integrate `f(x)=sec(x)/(4-3tan(x))`?

Answer 1

`int (sec(x))/(4-3tan(x))dx = int (1/cos(x))/(4-3tan(x)) dx`

Let's start with `t =tan(x/2)`

`sec(x) = (1+t^2)/(1-t^2)`

`tan(x) = (2t)/(1-t^2)`

`dx = 2/(1+t^2) dt`

So, the integral become :

`=> int((1+t^2)/(1-t^2))/(4-(6t)/(1-t^2))*2/(1+t^2)dt`

`=>int((1+t^2)/(1-t^2))/((4(1-t^2))/(1-t^2)-(6t)/(1-t^2))*2/(1+t^2)dt`

`=>int((1+t^2)/(1-t^2))/((-4t^2-6t+4)/(1-t^2))*2/(1+t^2)dt`

`=>int(1+t^2)/((-4t^2-6t+4))*2/(1+t^2)dt`

`=>int2/(2(-2t^2-3t+2))dt`

`=>int1/(-2t^2-3t+2)dt`

`Delta = b^2-4ac = 25`

`x_1 = 1/2`

`x_2 = -2`

Factorize :

`=>int-1/((2t-1)(t+2))dt`

Partial fraction :

`=>1/5int 1/(t+2)dt-1/5int2/(2t-1)dt`

`=>1/5[ln(|t+2|)]-1/5[ln(|2t-1|)]+C`

`=> 1/5((ln(|t+2|)-ln(|2t-1|))+C`

Substitute back for `t = tan(1/2x)`

`=> 1/5(ln(|tan(1/2x)+2|)-ln(|2tan(1/2x)-1|))+C`