integral of sin^9(x)cos^3(x)

Lösung

\int sin^{9} (x) cos^{3} (x) d x

- \frac{cos ^{4} ( x )}{4} + \frac{2 cos ^{6} ( x )}{3} - \frac{3 cos ^{8} ( x )}{4} + \frac{2 cos ^{10} ( x )}{5} - \frac{cos ^{12} ( x )}{12} + C

Schritte zur Lösung

\int sin^{9} (x) cos^{3} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende U-Substitution an

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

Multipliziere aus

- u^{3} (1 - u^{2})^{4} : - u^{3} + 4 u^{5} - 6 u^{7} + 4 u^{9} - u^{11}

= \int - u^{3} + 4 u^{5} - 6 u^{7} + 4 u^{9} - u^{11} 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 u^{3} d u + \int 4 u^{5} d u - \int 6 u^{7} d u + \int 4 u^{9} d u - \int u^{11} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int 4 u^{5} d u = \frac{2 u ^{6}}{3}

\int 6 u^{7} d u = \frac{3 u ^{8}}{4}

\int 4 u^{9} d u = \frac{2 u ^{10}}{5}

\int u^{11} d u = \frac{u ^{12}}{12}

= - \frac{u ^{4}}{4} + \frac{2 u ^{6}}{3} - \frac{3 u ^{8}}{4} + \frac{2 u ^{10}}{5} - \frac{u ^{12}}{12}

Setze in

u = cos (x)

ein

= - \frac{cos ^{4} ( x )}{4} + \frac{2 cos ^{6} ( x )}{3} - \frac{3 cos ^{8} ( x )}{4} + \frac{2 cos ^{10} ( x )}{5} - \frac{cos ^{12} ( x )}{12}

Füge eine Konstante zur Lösung hinzu

= - \frac{cos ^{4} ( x )}{4} + \frac{2 cos ^{6} ( x )}{3} - \frac{3 cos ^{8} ( x )}{4} + \frac{2 cos ^{10} ( x )}{5} - \frac{cos ^{12} ( x )}{12} + C

d er i v a t i v e y = \frac{7 x - 24}{x ^{2} + 2 x}

\int \frac{1}{3 x ^{6}} d x

\frac{d}{d x} (ln (\frac{e ^{x}}{x}))

\frac{d}{d x} (5 x^{6})

s \to - 1 lim (\frac{1}{s ^{2} + 2 s + 1})