integral of cos^6(3x)

Lösung

\int cos^{6} (3 x) d x

\frac{1}{3} (\frac{1}{6} cos^{5} (3 x) sin (3 x) + \frac{5}{6} (\frac{1}{4} cos^{3} (3 x) sin (3 x) + \frac{3}{8} (3 x + \frac{1}{2} sin (6 x)))) + C

Schritte zur Lösung

\int cos^{6} (3 x) d x

Wende U-Substitution an

= \int cos^{6} (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^{6} (u) d u

Wende integrale Reduktion an

= \frac{1}{3} (\frac{sin ( u ) cos ^{5} ( u )}{6} + \frac{5}{6} \cdot \int cos^{4} (u) d u)

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

= \frac{1}{3} (\frac{sin ( u ) cos ^{5} ( u )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (u) sin (u) + \frac{3}{8} (u + \frac{1}{2} sin (2 u))))

Setze in

u = 3 x

ein

= \frac{1}{3} (\frac{sin ( 3 x ) cos ^{5} ( 3 x )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (3 x) sin (3 x) + \frac{3}{8} (3 x + \frac{1}{2} sin (2 \cdot 3 x))))

Vereinfache

\frac{1}{3} (\frac{sin ( 3 x ) cos ^{5} ( 3 x )}{6} + \frac{5}{6} (\frac{1}{4} cos^{3} (3 x) sin (3 x) + \frac{3}{8} (3 x + \frac{1}{2} sin (2 \cdot 3 x)))) : \frac{1}{3} (\frac{1}{6} cos^{5} (3 x) sin (3 x) + \frac{5}{6} (\frac{1}{4} cos^{3} (3 x) sin (3 x) + \frac{3}{8} (3 x + \frac{1}{2} sin (6 x))))

= \frac{1}{3} (\frac{1}{6} cos^{5} (3 x) sin (3 x) + \frac{5}{6} (\frac{1}{4} cos^{3} (3 x) sin (3 x) + \frac{3}{8} (3 x + \frac{1}{2} sin (6 x))))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (\frac{1}{6} cos^{5} (3 x) sin (3 x) + \frac{5}{6} (\frac{1}{4} cos^{3} (3 x) sin (3 x) + \frac{3}{8} (3 x + \frac{1}{2} sin (6 x)))) + C

\int_{1}^{2} \frac{11}{x ^{3} + 2 x} d x

t an g e n t 3 x^{2} + 2 x - 10

in v erse l a pl a ce \frac{( 2 s + 1 )}{s ( s + 1 ) ^{2}}

p a r i t y y = x^{s i n (x)}

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = - 4, y^{^{'}} (0) = 0