integral of (tan(x))/(sec(x)+tan(x))

解答

\int \frac{tan ( x )}{sec ( x ) + tan ( x )} d x

\frac{2}{tan ( \frac{x}{2} ) + 1} + x + C

求解步骤

\int \frac{tan ( x )}{sec ( x ) + tan ( x )} d x

使用三角恒等式改写

= \int \frac{sin ( x )}{1 + sin ( x )} d x

\frac{sin ( x )}{1 + sin ( x )} = - \frac{1}{1 + sin ( x )} + 1

= \int - \frac{1}{1 + sin ( x )} + 1 d x

使用积分加法定则:

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

= - \int \frac{1}{1 + sin ( x )} d x + \int 1 d x

\int \frac{1}{1 + sin ( x )} d x = - \frac{2}{tan ( \frac{x}{2} ) + 1}

\int 1 d x = x

= - (- \frac{2}{tan ( \frac{x}{2} ) + 1}) + x

化简

= \frac{2}{tan ( \frac{x}{2} ) + 1} + x

解答补常数

= \frac{2}{tan ( \frac{x}{2} ) + 1} + x + C

\int sin^{2} (\frac{x}{5}) d x

\frac{d y}{d t} = y + 1 + 2 sin (t)

\frac{\partial}{\partial z} (x^{2} y cos (\frac{z}{t}))

\int 1 - (x - 1)^{2} d x

\frac{d}{d x} (x^{3} - 12 x + 1)