integral from 0 to pi/2 of 1-sin^2(x)

Lösung

\int_{0}^{\frac{π}{2}} 1 - sin^{2} (x) d x

\frac{π}{4}

Dezimale

0.78539 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} 1 - sin^{2} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

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

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

= \frac{1}{2} (\frac{π}{2} + 0)

Vereinfache

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

= \frac{π}{4}

\int_{0}^{2} 3 x d x

\int_{1}^{5} \frac{3 x}{x ^{3}} d x

\int_{0}^{1} x^{2} - x^{5} d x

\int_{0}^{1} π (4 - x^{2})^{2} d x

\int_{1}^{x} r t^{r - 1} d t