integral from 0 to 2pi of sin(x)sin(x+1)

Lösung

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

π cos (1) + \frac{- sin ( 4 π + 1 ) + sin ( 1 )}{4}

Dezimale

1.69740 \dots

Schritte zur Lösung

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

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

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

= \frac{1}{2} (- \frac{1}{2} (sin (4 π + 1) - sin (1)) + 2 π cos (1))

Vereinfache

\frac{1}{2} (- \frac{1}{2} (sin (4 π + 1) - sin (1)) + 2 π cos (1)) : π cos (1) + \frac{- sin ( 4 π + 1 ) + sin ( 1 )}{4}

= π cos (1) + \frac{- sin ( 4 π + 1 ) + sin ( 1 )}{4}

\int 2 v d v

d er i v a t i v e f (x) = lo g_{4} (x)

\frac{d x}{d y} = x - x^{2}

l a pl a ce t r an s f or m t^{2} e^{4 t}

\frac{d}{d x} (e^{a r c s i n (x)})