integral of (2-cot(x))(1-csc(x))

解

\int (2 - cot (x)) (1 - csc (x)) d x

2 x - 2 ln tan (\frac{x}{2}) - ln ∣ sin (x) ∣ - csc (x) + C

解答ステップ

\int (2 - cot (x)) (1 - csc (x)) d x

三角関数の公式を使用して書き換える

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

拡張

\frac{( 2 sin ( x ) - cos ( x ) ) ( sin ( x ) - 1 )}{sin ^{2} ( x )} : 2 - \frac{2}{sin ( x )} - \frac{cos ( x )}{sin ( x )} + \frac{cos ( x )}{sin ^{2} ( x )}

= \int 2 - \frac{2}{sin ( x )} - \frac{cos ( x )}{sin ( x )} + \frac{cos ( x )}{sin ^{2} ( 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 2 d x - \int \frac{2}{sin ( x )} d x - \int \frac{cos ( x )}{sin ( x )} d x + \int \frac{cos ( x )}{sin ^{2} ( x )} d x

\int 2 d x = 2 x

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

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

\int \frac{cos ( x )}{sin ^{2} ( x )} d x = - csc (x)

= 2 x - 2 ln tan (\frac{x}{2}) - ln ∣ sin (x) ∣ - csc (x)

定数を解答に追加する

= 2 x - 2 ln tan (\frac{x}{2}) - ln ∣ sin (x) ∣ - csc (x) + C