integral of (sin^3(x))/(2-sin^2(x))

Lösung

\int \frac{sin ^{3} ( x )}{2 - sin ^{2} ( x )} d x

- 2 arctan (cos (x)) + cos (x) + C

Schritte zur Lösung

\int \frac{sin ^{3} ( x )}{2 - sin ^{2} ( x )} d x

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

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

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

\int \frac{2}{u ^{2} + 1} d u = 2 arctan (u)

\int 1 d u = u

= - (2 arctan (u) - u)

Setze in

u = cos (x)

ein

= - (2 arctan (cos (x)) - cos (x))

Vereinfache

= - 2 arctan (cos (x)) + cos (x)

Füge eine Konstante zur Lösung hinzu

= - 2 arctan (cos (x)) + cos (x) + C

\int \frac{sec ( θ )}{sec ( θ ) - tan ( θ )} d θ

\int \frac{x ^{2} - 9 x + 26}{x ^{3} + x ^{2} + 4 x + 12} d x

\int \frac{3 e ^{a}}{4 - e ^{2 a}} d a

\int \frac{1}{2 - 4 x} d x

\int \frac{x ^{3}}{6 + x ^{2}} d x