integral from 0 to 2pi of sin^2(θ)

Lösung

\int_{0}^{2 π} sin^{2} (θ) d θ

π

Dezimale

3.14159 \dots

Schritte zur Lösung

\int_{0}^{2 π} sin^{2} (θ) d θ

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Entferne die Konstante:

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

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

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}^{2 π} 1 d θ - \int_{0}^{2 π} cos (2 θ) d θ)

\int_{0}^{2 π} 1 d θ = 2 π

\int_{0}^{2 π} cos (2 θ) d θ = 0

= \frac{1}{2} (2 π - 0)

Vereinfache

\frac{1}{2} (2 π - 0) : π

= π

t an g e n t f (x) = 2 x^{2} - 5 x

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