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integral of sqrt(tan(θ))

Lösung

\int tan (θ) d θ

22 (- \frac{1}{8} (ln 2 tan (θ) + 22 tan (θ) + 2 - 2 arctan (2 tan (θ) + 1)) + \frac{1}{8} (ln 2 tan (θ) - 22 tan (θ) + 2 + 2 arctan (2 tan (θ) - 1))) + C

Schritte zur Lösung

\int tan (θ) d θ

Trigonometrische Substitution anwenden

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

Wende integrale Substitution an

= \int \frac{2 v}{4 + v ^{2}} d v

Entferne die Konstante:

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

= 2 \cdot \int \frac{v}{4 + v ^{2}} d v

Wende U-Substitution an

= 2 \cdot \int \frac{2 w ^{2}}{4 + w ^{4}} d w

Entferne die Konstante:

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

= 2 \cdot 2 \cdot \int \frac{w ^{2}}{4 + w ^{4}} d w

Ermittle den Partialbruch von

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

= 2 \cdot 2 \cdot \int - \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 2 (- \int \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w + \int \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w)

\int \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w = \frac{1}{8} (ln w^{2} + 2 w + 2 - 2 arctan (w + 1))

\int \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w = \frac{1}{8} (ln w^{2} - 2 w + 2 + 2 arctan (w - 1))

= 2 \cdot 2 (- \frac{1}{8} (ln w^{2} + 2 w + 2 - 2 arctan (w + 1)) + \frac{1}{8} (ln w^{2} - 2 w + 2 + 2 arctan (w - 1)))

Ersetze zurück

= 2 \cdot 2 (- \frac{1}{8} (ln (2 tan (θ))^{2} + 2 2 tan (θ) + 2 - 2 arctan (2 tan (θ) + 1)) + \frac{1}{8} (ln (2 tan (θ))^{2} - 2 2 tan (θ) + 2 + 2 arctan (2 tan (θ) - 1)))

Vereinfache

2 \cdot 2 (- \frac{1}{8} (ln (2 tan (θ))^{2} + 2 2 tan (θ) + 2 - 2 arctan (2 tan (θ) + 1)) + \frac{1}{8} (ln (2 tan (θ))^{2} - 2 2 tan (θ) + 2 + 2 arctan (2 tan (θ) - 1))) : 22 (- \frac{1}{8} (ln 2 tan (θ) + 22 tan (θ) + 2 - 2 arctan (2 tan (θ) + 1)) + \frac{1}{8} (ln 2 tan (θ) - 22 tan (θ) + 2 + 2 arctan (2 tan (θ) - 1)))

= 22 (- \frac{1}{8} (ln 2 tan (θ) + 22 tan (θ) + 2 - 2 arctan (2 tan (θ) + 1)) + \frac{1}{8} (ln 2 tan (θ) - 22 tan (θ) + 2 + 2 arctan (2 tan (θ) - 1)))

Füge eine Konstante zur Lösung hinzu

= 22 (- \frac{1}{8} (ln 2 tan (θ) + 22 tan (θ) + 2 - 2 arctan (2 tan (θ) + 1)) + \frac{1}{8} (ln 2 tan (θ) - 22 tan (θ) + 2 + 2 arctan (2 tan (θ) - 1))) + C

\int 2019 + sec (x) tan (x) - e^{x} d x

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