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

解

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

- cot (\frac{x}{2}) + ln ∣ 1 - cos (x) ∣ + C

解答ステップ

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

拡張

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

\int \frac{1}{1 - cos ( x )} d x = - cot (\frac{x}{2})

\int \frac{sin ( x )}{1 - cos ( x )} d x = ln ∣ 1 - cos (x) ∣

= - cot (\frac{x}{2}) + ln ∣ 1 - cos (x) ∣

定数を解答に追加する

= - cot (\frac{x}{2}) + ln ∣ 1 - cos (x) ∣ + C