integral from 0 to pi/2 of cos(sqrt(2x))

Lösung

\int_{0}^{\frac{π}{2}} cos (2 x) d x

π sin (π) + cos (π) - 1

Dezimale

0.53624 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} cos (2 x) d x

Wende U-Substitution an

= \int_{0}^{π} cos (u) \frac{1}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{0}^{π} cos (u) d u

Wende U-Substitution an

= \frac{1}{2} \cdot \int_{0}^{π} cos (v) \cdot 2 v d v

Entferne die Konstante:

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

= \frac{1}{2} \cdot 2 \cdot \int_{0}^{π} cos (v) v d v

Wende die partielle Integration an

= \frac{1}{2} \cdot 2 [v sin (v) - \int sin (v) d v]_{0}^{π}

\int sin (v) d v = - cos (v)

= \frac{1}{2} \cdot 2 [v sin (v) - (- cos (v))]_{0}^{π}

Vereinfache

\frac{1}{2} \cdot 2 [v sin (v) - (- cos (v))]_{0}^{π} : [v sin (v) + cos (v)]_{0}^{π}

= [v sin (v) + cos (v)]_{0}^{π}

Berechne die Grenzen:

π sin (π) + cos (π) - 1

= π sin (π) + cos (π) - 1

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