integral from 2 to 4 of arcsec(sqrt(t))

Lösung

\int_{2}^{4} arcsec (t) d t

- 3 + \frac{5 π}{6} + 1

Dezimale

1.88594 \dots

Schritte zur Lösung

\int_{2}^{4} arcsec (t) d t

Wende die partielle Integration an

= [t arcsec (t) - \int \frac{1}{2 t - 1} d t]_{2}^{4}

\int \frac{1}{2 t - 1} d t = t - 1

= [t arcsec (t) - t - 1]_{2}^{4}

Berechne die Grenzen:

- 3 + \frac{5 π}{6} + 1

= - 3 + \frac{5 π}{6} + 1

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