integral from 0 to pi/4 of sqrt(x^2+1)

Lösung

\int_{0}^{\frac{π}{4}} x^{2} + 1 d x

\frac{π π ^{2} + 16}{32} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

Dezimale

0.85995 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{4}} x^{2} + 1 d x

Trigonometrische Substitution anwenden

= \int_{0}^{a r c t a n (\frac{π}{4})} sec^{3} (u) d u

Wende integrale Reduktion an

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{π}{4})} sec (u) d u

\int_{0}^{a r c t a n (\frac{π}{4})} sec (u) d u = ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

Vereinfache

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{π}{4})} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

Berechne die Grenzen:

\frac{π π ^{2} + 16}{32}

= \frac{π π ^{2} + 16}{32} + \frac{1}{2} ln (\frac{π}{4} + \frac{π ^{2}}{16} + 1)

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