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Beliebt Rechnen >

integral of 1/(1+tan(β))

Lösung

\int \frac{1}{1 + tan ( β )} d β

Lösung

\frac{1}{2} (- \frac{1}{2} ln (sec^{2} (β)) + arctan (tan (β))) + \frac{1}{2} ln ∣ tan (β) + 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{1 + tan ( β )} d β

Wende U-Substitution an

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

Ermittle den Partialbruch von

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

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

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

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

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

= \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 + arctan (u)) + \frac{1}{2} ln ∣ u + 1 ∣

Setze in

u = tan (β)

ein

= \frac{1}{2} (- \frac{1}{2} ln tan^{2} (β) + 1 + arctan (tan (β))) + \frac{1}{2} ln ∣ tan (β) + 1 ∣

\frac{1}{2} ln tan^{2} (β) + 1 = \frac{1}{2} ln (sec^{2} (β))

= \frac{1}{2} (- \frac{1}{2} ln (sec^{2} (β)) + arctan (tan (β))) + \frac{1}{2} ln ∣ tan (β) + 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (- \frac{1}{2} ln (sec^{2} (β)) + arctan (tan (β))) + \frac{1}{2} ln ∣ tan (β) + 1 ∣ + C

Graph

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