integral of cos^4(3x+2)

Lösung

\int cos^{4} (3 x + 2) d x

\frac{1}{3} cos^{3} (3 x + 2) sin (3 x + 2) + \frac{1}{32} (12 x + 8 - sin (4 (3 x + 2))) + C

Schritte zur Lösung

\int cos^{4} (3 x + 2) d x

Wende U-Substitution an

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

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int cos^{4} (u) d u

Vereinfache

cos^{4} (u) : cos^{3} (u) cos (u)

= \frac{1}{3} \cdot \int cos^{3} (u) cos (u) d u

Wende die partielle Integration an

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

\int - 3 cos^{2} (u) sin^{2} (u) d u = - \frac{3}{8} (u - \frac{1}{4} sin (4 u))

= \frac{1}{3} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Setze in

u = 3 x + 2

ein

= \frac{1}{3} (cos^{3} (3 x + 2) sin (3 x + 2) - (- \frac{3}{8} (3 x + 2 - \frac{1}{4} sin (4 (3 x + 2)))))

Vereinfache

\frac{1}{3} (cos^{3} (3 x + 2) sin (3 x + 2) - (- \frac{3}{8} (3 x + 2 - \frac{1}{4} sin (4 (3 x + 2))))) : \frac{1}{3} cos^{3} (3 x + 2) sin (3 x + 2) + \frac{1}{32} (12 x + 8 - sin (4 (3 x + 2)))

= \frac{1}{3} cos^{3} (3 x + 2) sin (3 x + 2) + \frac{1}{32} (12 x + 8 - sin (4 (3 x + 2)))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} cos^{3} (3 x + 2) sin (3 x + 2) + \frac{1}{32} (12 x + 8 - sin (4 (3 x + 2))) + C

\int r^{3} 4 - r^{2} d r

\int x (2 x - 1)^{100} d x

\int (5 x^{4} + 8 x + 7) d x

\int \frac{1}{2 z} d z

\int cos (2 x) - sin (2 x) d x