integral of 7(sin(x))^4(cos(x))^2

Lösung

\int 7 (sin (x))^{4} (cos (x))^{2} d x

7 (\frac{3}{8} x - \frac{1}{24} sin^{3} (x) cos (x) - \frac{3}{16} sin (2 x) + \frac{1}{6} sin^{5} (x) cos (x) - \frac{5}{32} (2 x - sin (2 x))) + C

Schritte zur Lösung

\int 7 (sin (x))^{4} (cos (x))^{2} d x

Entferne die Konstante:

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

= 7 \cdot \int (sin (x))^{4} (cos (x))^{2} d x

Vereinfache

= 7 \cdot \int sin^{4} (x) cos^{2} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 7 \cdot \int sin^{4} (x) (1 - sin^{2} (x)) d x

Multipliziere aus

sin^{4} (x) (1 - sin^{2} (x)) : sin^{4} (x) - sin^{6} (x)

= 7 \cdot \int sin^{4} (x) - sin^{6} (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

= 7 (\int sin^{4} (x) d x - \int sin^{6} (x) d x)

\int sin^{4} (x) d x = - \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))

\int sin^{6} (x) d x = - \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))

= 7 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)))))

Vereinfache

7 (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x)) - (- \frac{cos ( x ) sin ^{5} ( x )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (x) cos (x) + \frac{3}{8} (x - \frac{1}{2} sin (2 x))))) : 7 (\frac{3}{8} x - \frac{1}{24} sin^{3} (x) cos (x) - \frac{3}{16} sin (2 x) + \frac{1}{6} sin^{5} (x) cos (x) - \frac{5}{32} (2 x - sin (2 x)))

= 7 (\frac{3}{8} x - \frac{1}{24} sin^{3} (x) cos (x) - \frac{3}{16} sin (2 x) + \frac{1}{6} sin^{5} (x) cos (x) - \frac{5}{32} (2 x - sin (2 x)))

Füge eine Konstante zur Lösung hinzu

= 7 (\frac{3}{8} x - \frac{1}{24} sin^{3} (x) cos (x) - \frac{3}{16} sin (2 x) + \frac{1}{6} sin^{5} (x) cos (x) - \frac{5}{32} (2 x - sin (2 x))) + C

x \frac{d y}{d x} = x^{3} + 14 x^{3} y

\int c x y + \frac{1}{4}

t an g e n t f (x) = \frac{x}{4 x - 7}, at x = 1

t an g e n t - 5 x^{2} + x - 2

\int_{0}^{9} \frac{1}{81 + x ^{2}} d x