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

Lösung

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

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

Schritte zur Lösung

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

Vereinfache

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

Wende U-Substitution an

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

Ermittle den Partialbruch von

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

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

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

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

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

Setze in

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

ein

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

Füge eine Konstante zur Lösung hinzu

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

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