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

解

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

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

解答ステップ

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

三角関数の公式を使用して書き換える

= \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

総和規則を適用する:

\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

簡素化

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

定数を解答に追加する

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