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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

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

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

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

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

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

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

Setze in

u = tan (\frac{x}{2})

ein

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

Vereinfache

2 (- \frac{1}{2} ln tan (\frac{x}{2}) + 1 + \frac{1}{2} (\frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1 + arctan (tan (\frac{x}{2})))) : - ln tan (\frac{x}{2}) + 1 + \frac{1}{2} ln (sec^{2} (\frac{x}{2})) + arctan (tan (\frac{x}{2}))

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

Füge eine Konstante zur Lösung hinzu

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

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