integral of (1/(1-sin(x))-1/(1+sin(x)))

解

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

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

解答ステップ

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

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

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

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

簡素化

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

定数を解答に追加する

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