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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

Wende die Summenregel an:

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

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

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

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

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

Füge eine Konstante zur Lösung hinzu

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

\int y (a + b y^{2}) d y

\int - \frac{320}{x ^{6}} d x

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\int \frac{9 x}{x ^{2} + 3} d x

\int \frac{u ^{3}}{( u + 2 ) ^{2}} d u