integral from 0 to pi/2 of sin^2(2θ)

Lösung

\int_{0}^{\frac{π}{2}} sin^{2} (2 θ) d θ

\frac{π}{4}

Dezimale

0.78539 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} sin^{2} (2 θ) d θ

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{\frac{π}{2}} \frac{1}{2} (1 - cos (4 θ)) 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}^{\frac{π}{2}} 1 - cos (4 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}^{\frac{π}{2}} 1 d θ - \int_{0}^{\frac{π}{2}} cos (4 θ) d θ)

\int_{0}^{\frac{π}{2}} 1 d θ = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (4 θ) d θ = 0

= \frac{1}{2} (\frac{π}{2} - 0)

Vereinfache

\frac{1}{2} (\frac{π}{2} - 0) : \frac{π}{4}

= \frac{π}{4}

\int_{0}^{\frac{1}{2}} x e^{3 x} d x

\int_{0}^{y} x^{2} e^{- x y} d x

\int_{4}^{0} x d x

\int_{\frac{π}{6}}^{2 π} cos (x) d x

\int_{2}^{5} x - 1 d x