integral of sin(x)sqrt(1+cos^2(x))

Lösung

\int sin (x) 1 + cos^{2} (x) d x

- \frac{1}{2} cos (x) 1 + cos^{2} (x) - \frac{1}{2} ln cos (x) + 1 + cos^{2} (x) + C

Schritte zur Lösung

\int sin (x) 1 + cos^{2} (x) d x

Wende U-Substitution an

= \int - 1 + u^{2} d u

Entferne die Konstante:

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

= - \int 1 + u^{2} d u

Trigonometrische Substitution anwenden

= - \int sec^{3} (v) d v

Wende integrale Reduktion an

= - (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

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

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

Ersetze zurück

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

Vereinfache

- (\frac{sec ^{2} ( arctan ( cos ( x ) ) ) sin ( arctan ( cos ( x ) ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (cos (x))) + sec (arctan (cos (x))) ∣) : - \frac{1}{2} cos (x) 1 + cos^{2} (x) - \frac{1}{2} ln cos (x) + 1 + cos^{2} (x)

= - \frac{1}{2} cos (x) 1 + cos^{2} (x) - \frac{1}{2} ln cos (x) + 1 + cos^{2} (x)

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} cos (x) 1 + cos^{2} (x) - \frac{1}{2} ln cos (x) + 1 + cos^{2} (x) + C

\frac{d P}{d t} = P (1 0^{- 1} - P \cdot 1 0^{- 7}), P (0) = 7000

d er i v a t i v e f (x) = x^{- \frac{1}{2}}

\frac{d s}{d t}

d er i v a t i v e h (x) = 5 sin^{3} (3 + x)

x \to 2 lim (- x^{2} + 6 x - 7)