integral of cos^4(7x+7)

Lösung

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

\frac{1}{7} (cos^{3} (7 x + 7) sin (7 x + 7) + \frac{3}{8} (7 x + 7 - \frac{1}{4} sin (4 (7 x + 7)))) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int cos^{4} (u) \frac{1}{7} d u

Entferne die Konstante:

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

= \frac{1}{7} \cdot \int cos^{4} (u) d u

Vereinfache

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

= \frac{1}{7} \cdot \int cos^{3} (u) cos (u) d u

Wende die partielle Integration an

= \frac{1}{7} (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}{7} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Setze in

u = 7 x + 7

ein

= \frac{1}{7} (cos^{3} (7 x + 7) sin (7 x + 7) - (- \frac{3}{8} (7 x + 7 - \frac{1}{4} sin (4 (7 x + 7)))))

Vereinfache

= \frac{1}{7} (cos^{3} (7 x + 7) sin (7 x + 7) + \frac{3}{8} (7 x + 7 - \frac{1}{4} sin (4 (7 x + 7))))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{7} (cos^{3} (7 x + 7) sin (7 x + 7) + \frac{3}{8} (7 x + 7 - \frac{1}{4} sin (4 (7 x + 7)))) + C

d er i v a t i v e y = 5 y^{2} (4 y^{2} + 7)^{\frac{6}{5}}

\frac{d y}{d x} = \frac{7 x}{3 y}

t an g e n t 4 x^{2} + 2 x

t an g e n t y = 5 x - x^{2}

\frac{d}{d x} (\frac{ln ( 4 x )}{x})