Question

How do you integrate `sec(x)/(4-3tan(x)) dx`?

Answer 1

`1/5ln(tan(x/2)+2)-1/5Ln(2tan(x/2)-1)+C`

Explanation:

`int secx/(4-3tanx)*dx`

=`int dx/(4cosx-3sinx)`

After using `y=tan(x/2)`, `dx=(2dy)/(y^2+1)`, `cosx=(1-y^2)/(y^2+1)` and `sinx=(2y)/(y^2+1)` transforms, this integral became

`int ((2dy)/(y^2+1))/(4*(1-y^2)/(y^2+1)-3*(2y)/(y^2+1))`

=`int ((2dy)/(y^2+1))/((4-6y-4y^2)/(y^2^1))`

=-`int dy/(2y^2+3y-2)`

=-`int dy/((y+2)*(2y-1))`

=-`1/5int (5dy)/((y+2)*(2y-1))`

=-`1/5int ((2y+4)*dy)/((y+2)(2y-1))`+`1/5int ((2y-1)*dy)/((y+2)(2y-1))`

=`1/5int dy/(y+2)-2/5 int dy/(2y-1)`

=`1/5ln(y+2)-1/5Ln(2y-1)+C`

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