integral of (sin^3(x))/(1+cos^3(x))

Lösung

\int \frac{sin ^{3} ( x )}{1 + cos ^{3} ( x )} d x

\frac{1}{2} ln 4 cos^{2} (x) - 4 cos (x) + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 cos (x) - 1)) + C

Schritte zur Lösung

\int \frac{sin ^{3} ( x )}{1 + cos ^{3} ( x )} d x

Wende U-Substitution an

= \int \frac{u - 1}{u ^{2} - u + 1} d u

Vervollständige das Quadrat

u^{2} - u + 1 : (u - \frac{1}{2})^{2} + \frac{3}{4}

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

Wende U-Substitution an

= \int \frac{2 ( 2 v - 1 )}{4 v ^{2} + 3} d v

Entferne die Konstante:

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

= 2 \cdot \int \frac{2 v - 1}{4 v ^{2} + 3} d v

Multipliziere aus

\frac{2 v - 1}{4 v ^{2} + 3} : \frac{2 v}{4 v ^{2} + 3} - \frac{1}{4 v ^{2} + 3}

Wende die Summenregel an:

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

= 2 (\int \frac{2 v}{4 v ^{2} + 3} d v - \int \frac{1}{4 v ^{2} + 3} d v)

\int \frac{2 v}{4 v ^{2} + 3} d v = \frac{1}{4} ln 4 v^{2} + 3

\int \frac{1}{4 v ^{2} + 3} d v = \frac{1}{2 3} arctan (\frac{2}{3} v)

= 2 (\frac{1}{4} ln 4 v^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} v))

Ersetze zurück

= 2 (\frac{1}{4} ln 4 (cos (x) - \frac{1}{2})^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} (cos (x) - \frac{1}{2})))

Vereinfache

2 (\frac{1}{4} ln 4 (cos (x) - \frac{1}{2})^{2} + 3 - \frac{1}{2 3} arctan (\frac{2}{3} (cos (x) - \frac{1}{2}))) : \frac{1}{2} ln 4 cos^{2} (x) - 4 cos (x) + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 cos (x) - 1))

= \frac{1}{2} ln 4 cos^{2} (x) - 4 cos (x) + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 cos (x) - 1))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} ln 4 cos^{2} (x) - 4 cos (x) + 4 - \frac{1}{3} arctan (\frac{1}{3} (2 cos (x) - 1)) + C

\int 9 cot^{5} (x) d x

\int \frac{tan ( x )}{1 + tan ^{2} ( x )} d x

\int 1 + sin (y) d y

\int \frac{1}{x x - a ^{2}} d x

\int - 8 sin (7 x) d x