integral of cos^4(5x+8)

Lösung

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

\frac{1}{160} (32 cos^{3} (5 x + 8) sin (5 x + 8) + 3 (20 x + 32 - sin (4 (5 x + 8)))) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int cos^{4} (u) \frac{1}{5} d u

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int cos^{4} (u) d u

Vereinfache

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

= \frac{1}{5} \cdot \int cos^{3} (u) cos (u) d u

Wende die partielle Integration an

= \frac{1}{5} (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}{5} (cos^{3} (u) sin (u) - (- \frac{3}{8} (u - \frac{1}{4} sin (4 u))))

Setze in

u = 5 x + 8

ein

= \frac{1}{5} (cos^{3} (5 x + 8) sin (5 x + 8) - (- \frac{3}{8} (5 x + 8 - \frac{1}{4} sin (4 (5 x + 8)))))

Vereinfache

\frac{1}{5} (cos^{3} (5 x + 8) sin (5 x + 8) - (- \frac{3}{8} (5 x + 8 - \frac{1}{4} sin (4 (5 x + 8))))) : \frac{1}{160} (32 cos^{3} (5 x + 8) sin (5 x + 8) + 3 (20 x + 32 - sin (4 (5 x + 8))))

= \frac{1}{160} (32 cos^{3} (5 x + 8) sin (5 x + 8) + 3 (20 x + 32 - sin (4 (5 x + 8))))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{160} (32 cos^{3} (5 x + 8) sin (5 x + 8) + 3 (20 x + 32 - sin (4 (5 x + 8)))) + C

n = 1 \sum \infty \frac{n}{3 ^{n}}

\frac{\partial}{\partial θ} (ln (θ + 1))

\int \frac{( e ^{t a n (x)} )}{cos ^{2} ( x )} d x

y^{^{''}} + 5 y^{^{'}} + 2 y = 7

t an g e n t y = 2 x^{3}, (2, 16)