integral of sqrt(1+4θ^2)

Lösung

\int 1 + 4 θ^{2} d θ

\frac{1}{4} (2 θ 4 θ^{2} + 1 + ln 2 θ + 1 + 4 θ^{2}) + C

Schritte zur Lösung

\int 1 + 4 θ^{2} d θ

Trigonometrische Substitution anwenden

= \int \frac{1}{2} sec^{3} (u) d u

Entferne die Konstante:

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

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

Wende integrale Reduktion an

= \frac{1}{2} (\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) ∣

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

Setze in

u = arctan (2 θ)

ein

= \frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 θ ) ) sin ( arctan ( 2 θ ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 θ)) + sec (arctan (2 θ)) ∣)

Vereinfache

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 θ ) ) sin ( arctan ( 2 θ ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 θ)) + sec (arctan (2 θ)) ∣) : \frac{1}{4} (2 θ 4 θ^{2} + 1 + ln 2 θ + 1 + 4 θ^{2})

= \frac{1}{4} (2 θ 4 θ^{2} + 1 + ln 2 θ + 1 + 4 θ^{2})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} (2 θ 4 θ^{2} + 1 + ln 2 θ + 1 + 4 θ^{2}) + C

l a pl a ce t r an s f or m e^{2 t} (t - 1)^{2}

\frac{d}{d x} ((1 - 3 x) arccoth (3 x))

y^{^{'}} + 4 x y^{3} = 0

\frac{d}{d x} (\frac{4 x ^{3}}{3} + cos (x) + 1)

\frac{d}{d x} (sin (n x))