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

Lösung

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Multipliziere aus

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

= \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

= \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)))

= - \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

- \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)))) : \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))

= \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

= \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

\int e^{- s x} \cdot x^{2} d x

t an g e n t - x^{2} + 9 x + 3, at x = 5

n = 2 \sum \infty e^{3 - 4 n}

d er i v a t i v e f (x) = (2 x - 8) 4 (x^{2} + x + 1) 5

in v erse l a pl a ce \frac{s + 2}{s ^{2} - s - 12}