integral of 8sec^3(3x)

Lösung

\int 8 sec^{3} (3 x) d x

\frac{8}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣) + C

Schritte zur Lösung

\int 8 sec^{3} (3 x) d x

Entferne die Konstante:

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

= 8 \cdot \int sec^{3} (3 x) d x

Wende U-Substitution an

= 8 \cdot \int sec^{3} (u) \frac{1}{3} d u

Entferne die Konstante:

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

= 8 \cdot \frac{1}{3} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 8 \cdot \frac{1}{3} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = 3 x

ein

= 8 \cdot \frac{1}{3} (\frac{sec ^{2} ( 3 x ) sin ( 3 x )}{2} + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣)

Vereinfache

8 \cdot \frac{1}{3} (\frac{sec ^{2} ( 3 x ) sin ( 3 x )}{2} + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣) : \frac{8}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣)

= \frac{8}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣)

Füge eine Konstante zur Lösung hinzu

= \frac{8}{3} (\frac{1}{2} sec^{2} (3 x) sin (3 x) + \frac{1}{2} ln ∣ tan (3 x) + sec (3 x) ∣) + C

y^{^{'}} + y \cdot cos (t) = sin (t) \cdot cos (t)

e x p an d x^{3} (4 - x)

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 2

\int 3 sin (x) cos (x) d x

y^{^{''}} + 9 y = \frac{1}{sin ( 3 t )}