integral from 0 to pi/6 of tan(x)

Lösung

\int_{0}^{\frac{π}{6}} tan (x) d x

- \frac{1}{2} ln (3) + ln (2)

Dezimale

0.14384 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{6}} tan (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{\frac{π}{6}} \frac{sin ( x )}{cos ( x )} d x

Wende U-Substitution an

= \int_{1}^{\frac{3}{2}} - \frac{1}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{3}{2}}^{1} - \frac{1}{u} d u

Entferne die Konstante:

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

= - (- \int_{\frac{3}{2}}^{1} \frac{1}{u} d u)

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{\frac{3}{2}}^{1})

Vereinfache

= [ln ∣ u ∣]_{\frac{3}{2}}^{1}

Berechne die Grenzen:

- (\frac{1}{2} ln (3) - ln (2))

= - (\frac{1}{2} ln (3) - ln (2))

Vereinfache

= - \frac{1}{2} ln (3) + ln (2)

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