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

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Answer 1 · 2018-03-28T14:38:02

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

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

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

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

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

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

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

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

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

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

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

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

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

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