integral from 0 to pi of sin^4(at)

Lösung

\int_{0}^{π} sin^{4} (a t) d t

\frac{- \frac{1}{4} sin ^{3} ( πa ) cos ( πa ) + \frac{3}{8} ( πa - \frac{1}{2} sin ( 2 πa ) )}{a}

Schritte zur Lösung

\int_{0}^{π} sin^{4} (a t) d t

Wende U-Substitution an

= \int_{0}^{aπ} \frac{sin ^{4} ( u )}{a} d u

Entferne die Konstante:

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

= \frac{1}{a} \cdot \int_{0}^{aπ} sin^{4} (u) d u

Wende integrale Reduktion an

= \frac{1}{a} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{aπ} + \frac{3}{4} \cdot \int_{0}^{aπ} sin^{2} (u) d u)

\int_{0}^{aπ} sin^{2} (u) d u = \frac{1}{2} (πa - \frac{1}{2} sin (2 πa))

= \frac{1}{a} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{aπ} + \frac{3}{4} \cdot \frac{1}{2} (πa - \frac{1}{2} sin (2 πa)))

Vereinfache

\frac{1}{a} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{aπ} + \frac{3}{4} \cdot \frac{1}{2} (πa - \frac{1}{2} sin (2 πa))) : \frac{1}{a} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{πa} + \frac{3}{8} (πa - \frac{1}{2} sin (2 πa)))

= \frac{1}{a} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{πa} + \frac{3}{8} (πa - \frac{1}{2} sin (2 πa)))

Berechne die Grenzen:

\frac{1}{4} sin^{3} (πa) cos (πa)

= \frac{1}{a} (- \frac{1}{4} sin^{3} (πa) cos (πa) + \frac{3}{8} (πa - \frac{1}{2} sin (2 πa)))

Vereinfache

= \frac{- \frac{1}{4} sin ^{3} ( πa ) cos ( πa ) + \frac{3}{8} ( πa - \frac{1}{2} sin ( 2 πa ) )}{a}

\int_{\frac{π}{2}}^{π} \frac{1}{2} e^{- \frac{θ}{4}} d θ

\int_{1}^{3} 12 x^{3} d x

\int_{0}^{3} \frac{x ^{4}}{81} d x

\int_{1}^{3} ln (t + 1) sin (t) d t

\int_{1}^{3} 2 e^{5 x} d x