integral of (sin(2x)-sin(3x))^2

Lösung

\int (sin (2 x) - sin (3 x))^{2} d x

x - \frac{1}{8} sin (4 x) - \frac{1}{12} sin (6 x) + \frac{1}{5} sin (5 x) - sin (x) + C

Schritte zur Lösung

\int (sin (2 x) - sin (3 x))^{2} d x

Wende U-Substitution an

= \int \frac{1}{2} sin^{2} (u) - sin (\frac{3 u}{2}) sin (u) + \frac{1}{2} sin^{2} (\frac{3 u}{2}) d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2} sin^{2} (u) d u - \int sin (\frac{3 u}{2}) sin (u) d u + \int \frac{1}{2} sin^{2} (\frac{3 u}{2}) d u

\int \frac{1}{2} sin^{2} (u) d u = \frac{1}{4} (u - \frac{1}{2} sin (2 u))

\int sin (\frac{3 u}{2}) sin (u) d u = - \frac{1}{5} sin (\frac{5 u}{2}) + sin (\frac{u}{2})

\int \frac{1}{2} sin^{2} (\frac{3 u}{2}) d u = \frac{1}{4} (u - \frac{1}{3} sin (3 u))

= \frac{1}{4} (u - \frac{1}{2} sin (2 u)) - (- \frac{1}{5} sin (\frac{5 u}{2}) + sin (\frac{u}{2})) + \frac{1}{4} (u - \frac{1}{3} sin (3 u))

Setze in

u = 2 x

ein

= \frac{1}{4} (2 x - \frac{1}{2} sin (2 \cdot 2 x)) - (- \frac{1}{5} sin (\frac{5 \cdot 2 x}{2}) + sin (\frac{2 x}{2})) + \frac{1}{4} (2 x - \frac{1}{3} sin (3 \cdot 2 x))

Vereinfache

\frac{1}{4} (2 x - \frac{1}{2} sin (2 \cdot 2 x)) - (- \frac{1}{5} sin (\frac{5 \cdot 2 x}{2}) + sin (\frac{2 x}{2})) + \frac{1}{4} (2 x - \frac{1}{3} sin (3 \cdot 2 x)) : x - \frac{1}{8} sin (4 x) - \frac{1}{12} sin (6 x) + \frac{1}{5} sin (5 x) - sin (x)

= x - \frac{1}{8} sin (4 x) - \frac{1}{12} sin (6 x) + \frac{1}{5} sin (5 x) - sin (x)

Füge eine Konstante zur Lösung hinzu

= x - \frac{1}{8} sin (4 x) - \frac{1}{12} sin (6 x) + \frac{1}{5} sin (5 x) - sin (x) + C

t an g e n t f (x) = \frac{5}{x}, (3, \frac{5}{3})

\frac{\partial}{\partial t} (e^{- 2 x} cos (9 π t))

t a y l or \frac{1}{2} arctan (2 x)

n = 1 \sum \infty 3 e^{- 4 n}

t an g e n t f (x) = \frac{1}{x ^{2} + 4}, (1, \frac{1}{5})