integral from 0 to pi/2 of sin^4(t)

Lösung

\int_{0}^{\frac{π}{2}} sin^{4} (t) d t

\frac{3 π}{16}

Dezimale

0.58904 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} sin^{4} (t) d t

Wende integrale Reduktion an

= - [\frac{cos ( t ) sin ^{3} ( t )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (t) d t

\int_{0}^{\frac{π}{2}} sin^{2} (t) d t = \frac{π}{4}

= - [\frac{cos ( t ) sin ^{3} ( t )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4}

Vereinfache

- [\frac{cos ( t ) sin ^{3} ( t )}{4}]_{0}^{\frac{π}{2}} + \frac{3}{4} \cdot \frac{π}{4} : - [\frac{1}{4} sin^{3} (t) cos (t)]_{0}^{\frac{π}{2}} + \frac{3 π}{16}

= - [\frac{1}{4} sin^{3} (t) cos (t)]_{0}^{\frac{π}{2}} + \frac{3 π}{16}

Berechne die Grenzen:

0

= - 0 + \frac{3 π}{16}

Vereinfache

= \frac{3 π}{16}

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

\int_{0}^{3} \frac{1}{1 + x + 1} d x

\int_{0}^{2} 9 y^{2} y^{3} + 1 d y

\int_{1}^{2} \frac{1}{x} d x

\int_{0}^{3} (x - 3) e^{- x} d x