integral of (sin(θ))/(sqrt(4+cos^2(θ)))

Lösung

\int \frac{sin ( θ )}{4 + cos ^{2} ( θ )} d θ

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

Schritte zur Lösung

\int \frac{sin ( θ )}{4 + cos ^{2} ( θ )} d θ

Wende U-Substitution an

= \int - \frac{1}{4 + u ^{2}} d u

Entferne die Konstante:

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

= - \int \frac{1}{4 + u ^{2}} d u

Trigonometrische Substitution anwenden

= - \int sec (v) d v

Nutze das gemeinsame Integral :

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

= - ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= - ln tan (arctan (\frac{1}{2} cos (θ))) + sec (arctan (\frac{1}{2} cos (θ)))

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{2 x ^{2} + x + 9}{( x + 1 ) ( x ^{2} + 4 )} d x

\int \frac{1}{x + x + 1} d x

\int (sin (3 x))^{7} d x

\int 3 x x d x

\int 1 - 4^{x} 2^{x} d x