integral from 0 to pi of sin^2(nx)

Lösung

\int_{0}^{π} sin^{2} (n x) d x

\frac{1}{2} (π - \frac{sin ( 2 πn )}{2 n})

Schritte zur Lösung

\int_{0}^{π} sin^{2} (n x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

\int_{0}^{π} 1 d x = π

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

= \frac{1}{2} (π - \frac{sin ( 2 πn )}{2 n})

\int 3 (3 x + 9)^{7} d x

n or ma l f (x) = lo g_{2} (2 x^{2} + 14), at x = 5

\frac{d x}{d t} = 2 + 2 x + t + t x

\frac{\partial}{\partial x} (x^{3} + y^{2})

\frac{\partial}{\partial x} (\frac{x - 2 y ^{2}}{x ^{2} - y ^{3}})