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

Lösung

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

\frac{1}{4} ln ∣ 1 + sin (x) ∣ - \frac{1}{2 ( 1 + sin ( x ) )} - \frac{1}{4} ln ∣ sin (x) - 1 ∣ + C

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

= \int \frac{1}{( 1 + u ) ( 1 - u ^{2} )} d u

Wende U-Substitution an

= \int \frac{1}{v ( - v ^{2} + 2 v )} d v

Faktorisiere

\frac{1}{v ( - v ^{2} + 2 v )}

= \int \frac{1}{- v ^{2} ( v - 2 )} d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{v ^{2} ( v - 2 )} d v

Ermittle den Partialbruch von

\frac{1}{v ^{2} ( v - 2 )} : - \frac{1}{4 v} - \frac{1}{2 v ^{2}} + \frac{1}{4 ( v - 2 )}

= - \int - \frac{1}{4 v} - \frac{1}{2 v ^{2}} + \frac{1}{4 ( v - 2 )} d v

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}{4 v} d v - \int \frac{1}{2 v ^{2}} d v + \int \frac{1}{4 ( v - 2 )} d v)

\int \frac{1}{4 v} d v = \frac{1}{4} ln ∣ v ∣

\int \frac{1}{2 v ^{2}} d v = - \frac{1}{2 v}

\int \frac{1}{4 ( v - 2 )} d v = \frac{1}{4} ln ∣ v - 2 ∣

= - (- \frac{1}{4} ln ∣ v ∣ - (- \frac{1}{2 v}) + \frac{1}{4} ln ∣ v - 2 ∣)

Ersetze zurück

= - (- \frac{1}{4} ln ∣ 1 + sin (x) ∣ - (- \frac{1}{2 ( 1 + sin ( x ) )}) + \frac{1}{4} ln ∣ 1 + sin (x) - 2 ∣)

Vereinfache

- (- \frac{1}{4} ln ∣ 1 + sin (x) ∣ - (- \frac{1}{2 ( 1 + sin ( x ) )}) + \frac{1}{4} ln ∣ 1 + sin (x) - 2 ∣) : \frac{1}{4} ln ∣ 1 + sin (x) ∣ - \frac{1}{2 ( 1 + sin ( x ) )} - \frac{1}{4} ln ∣ sin (x) - 1 ∣

= \frac{1}{4} ln ∣ 1 + sin (x) ∣ - \frac{1}{2 ( 1 + sin ( x ) )} - \frac{1}{4} ln ∣ sin (x) - 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln ∣ 1 + sin (x) ∣ - \frac{1}{2 ( 1 + sin ( x ) )} - \frac{1}{4} ln ∣ sin (x) - 1 ∣ + C

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

\int \frac{1}{2 x ^{2} + 8} d x

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

\int \frac{1}{x ( x - 2 ) ^{3}} d x

\int \frac{x}{4 x - 2} d x