integral of (cos(x))/(sin^3(x)-4sin(x))

Lösung

\int \frac{cos ( x )}{sin ^{3} ( x ) - 4 sin ( x )} d x

- \frac{1}{4} ln ∣ sin (x) ∣ + \frac{1}{8} ln ∣ sin (x) + 2 ∣ + \frac{1}{8} ln ∣ sin (x) - 2 ∣ + C

Schritte zur Lösung

\int \frac{cos ( x )}{sin ^{3} ( x ) - 4 sin ( x )} d x

Wende U-Substitution an

= \int \frac{1}{u ^{3} - 4 u} d u

Faktorisiere

u^{3} - 4 u : - (- u^{3} + 4 u)

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

\frac{1}{- u ^{3} + 4 u} : \frac{1}{4 u} - \frac{1}{8 ( u + 2 )} - \frac{1}{8 ( u - 2 )}

= - \int \frac{1}{4 u} - \frac{1}{8 ( u + 2 )} - \frac{1}{8 ( u - 2 )} 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{1}{4 u} d u - \int \frac{1}{8 ( u + 2 )} d u - \int \frac{1}{8 ( u - 2 )} d u)

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

\int \frac{1}{8 ( u + 2 )} d u = \frac{1}{8} ln ∣ u + 2 ∣

\int \frac{1}{8 ( u - 2 )} d u = \frac{1}{8} ln ∣ u - 2 ∣

= - (\frac{1}{4} ln ∣ u ∣ - \frac{1}{8} ln ∣ u + 2 ∣ - \frac{1}{8} ln ∣ u - 2 ∣)

Setze in

u = sin (x)

ein

= - (\frac{1}{4} ln ∣ sin (x) ∣ - \frac{1}{8} ln ∣ sin (x) + 2 ∣ - \frac{1}{8} ln ∣ sin (x) - 2 ∣)

Vereinfache

= - \frac{1}{4} ln ∣ sin (x) ∣ + \frac{1}{8} ln ∣ sin (x) + 2 ∣ + \frac{1}{8} ln ∣ sin (x) - 2 ∣

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{4} ln ∣ sin (x) ∣ + \frac{1}{8} ln ∣ sin (x) + 2 ∣ + \frac{1}{8} ln ∣ sin (x) - 2 ∣ + C

\int (csc^{2} (θ) + 9) d θ

a re a f (x) = cos (x), 0, \frac{π}{4}

t an g e n t f (x) = \frac{x ^{3}}{2} - \frac{4}{3 x}, at x = 2

y = e^{m x} m

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