f^'(x)=2sec(x)+3tan(x)-tan(pi/3)

解答

f^{^{'}} (x) = 2 sec (x) + 3 tan (x) - tan (\frac{π}{3})

f (x) = 2 ln (tan (x) + sec (x)) - 3 ln (cos (x)) - 3 x + c_{1}

求解步骤

f^{'} (x) = 2 sec (x) + 3 tan (x) - tan (\frac{π}{3})

若

f^{^{'}} (x) = g (x)

则

f (x) = \int g (x) d x

f (x) = \int 2 sec (x) + 3 tan (x) - tan (\frac{π}{3}) d x

\int 2 sec (x) + 3 tan (x) - tan (\frac{π}{3}) d x = 2 ln (tan (x) + sec (x)) - 3 ln (cos (x)) - 3 x + c_{1}

f (x) = 2 ln (tan (x) + sec (x)) - 3 ln (cos (x)) - 3 x + c_{1}

y^{^{''}} + (\frac{1}{x}) y^{^{'}} = 0

y^{^{'}} = \frac{- ( x y + 1 )}{x y + 1}

y^{^{'}} = (x - 3) (y + 1)^{\frac{2}{3}}

\int_{0}^{\frac{π}{2}} - 8 sin^{2} (3 x) d x

\frac{d y}{d x} = cos (x)