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

Lösung

integral

sin^{3} (x) cos^{3} (x)

- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6} + C

Schritte zur Lösung

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende U-Substitution an

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

Multipliziere aus

- u^{3} (1 - u^{2}) : - u^{3} + u^{5}

= \int - u^{3} + u^{5} 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 u^{5} d u

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

\int u^{5} d u = \frac{u ^{6}}{6}

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{6}

Setze in

u = cos (x)

ein

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}

Füge eine Konstante zur Lösung hinzu

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6} + C

f (x) = (\frac{( x - 1 )}{( x + 1 )})^{\frac{1}{2}}

in t e g r a l sin^{2} (x) cos^{3} (x)

in t e g r a l \frac{1}{v ^{4}}

in t e g r a l \frac{( x ^{3} + 2 )}{( 4 - x ^{2} )}

in t e g r a l x^{2} (x - 4)^{\frac{1}{2}}