integral from 0 to pi/4 of sec^4(x/4)

Lösung

\int_{0}^{\frac{π}{4}} sec^{4} (\frac{x}{4}) d x

4 - 4 2 + 2 - 22 2 + 2 + 7 + 42 + \frac{( - 4 2 + 2 - 2 2 2 + 2 + 7 + 4 2 ) ^{\frac{3}{2}}}{3}

Dezimale

0.80614 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{4}} sec^{4} (\frac{x}{4}) d x

Wende U-Substitution an

= \int_{0}^{\frac{π}{16}} sec^{4} (u) \cdot 4 d u

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{\frac{π}{16}} sec^{4} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 4 \cdot \int_{0}^{\frac{π}{16}} (1 + tan^{2} (u)) sec^{2} (u) d u

Wende U-Substitution an

= 4 \cdot \int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} 1 + v^{2} d v

Wende die Summenregel an:

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

= 4 \int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} 1 d v + \int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} v^{2} d v

\int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} 1 d v = - 4 2 + 2 - 22 2 + 2 + 7 + 42

\int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} v^{2} d v = \frac{( - 4 2 + 2 - 2 2 2 + 2 + 7 + 4 2 ) ^{\frac{3}{2}}}{3}

= 4 - 4 2 + 2 - 22 2 + 2 + 7 + 42 + \frac{( - 4 2 + 2 - 2 2 2 + 2 + 7 + 4 2 ) ^{\frac{3}{2}}}{3}

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