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

Lösung

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

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

Schritte zur Lösung

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

Multipliziere aus

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{1}{1 + 3 - x} d x

\int sin ((x^{2})^{2}) d (x^{2})^{2}

\int 18 x - x^{2} d x

\int \frac{x ^{3}}{16 - x ^{2}} d x

\int sin (ln (y)) d y