integral of sin(1-2x)cos(1+2x)

Lösung

\int sin (1 - 2 x) cos (1 + 2 x) d x

\frac{1}{2} (sin (2) x + \frac{1}{4} cos (4 x)) + C

Schritte zur Lösung

\int sin (1 - 2 x) cos (1 + 2 x) d x

Verwende die folgenden Identitäten:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 1 - 2 x + 1 + 2 x ) + sin ( 1 - 2 x - ( 1 + 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 sin (1 - 2 x + 1 + 2 x) + sin (1 - 2 x - (1 + 2 x)) d x

Vereinfache

sin (1 - 2 x + 1 + 2 x) + sin (1 - 2 x - (1 + 2 x)) : sin (2) - sin (4 x)

= \frac{1}{2} \cdot \int sin (2) - sin (4 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 sin (2) d x - \int sin (4 x) d x)

\int sin (2) d x = sin (2) x

\int sin (4 x) d x = - \frac{1}{4} cos (4 x)

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} (sin (2) x + \frac{1}{4} cos (4 x)) + C

\frac{d}{d x} ((8 x^{2} - 5 x) (13 x^{2} + 4))

\frac{\partial}{\partial x} (c e^{- x} sin (4 x))

\int 3 t e^{t} d t

\int \frac{x ^{2}}{x ^{3} + 3} d x

\int (sin (x))^{2} d x