integral from 0 to 1 of cosh^2(x)

Lösung

\int_{0}^{1} cosh^{2} (x) d x

\frac{4 e ^{2} + e ^{4} - 1}{8 e ^{2}}

Dezimale

1.40671 \dots

Schritte zur Lösung

\int_{0}^{1} cosh^{2} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{1} \frac{1 + cosh ( 2 x )}{2} d x

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{0}^{1} 1 + cosh (2 x) d x

Wende die Summenregel an:

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

= \frac{1}{2} (\int_{0}^{1} 1 d x + \int_{0}^{1} cosh (2 x) d x)

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} cosh (2 x) d x = \frac{e ^{4} - 1}{4 e ^{2}}

= \frac{1}{2} (1 + \frac{e ^{4} - 1}{4 e ^{2}})

Vereinfache

\frac{1}{2} (1 + \frac{e ^{4} - 1}{4 e ^{2}}) : \frac{4 e ^{2} + e ^{4} - 1}{8 e ^{2}}

= \frac{4 e ^{2} + e ^{4} - 1}{8 e ^{2}}

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