integral of 1/(1-tan^2(x))

Lösung

\int \frac{1}{1 - tan ^{2} ( x )} d x

\frac{1}{2} arctan (tan (x)) + \frac{1}{4} ln ∣ tan (x) + 1 ∣ - \frac{1}{4} ln ∣ tan (x) - 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{1 - tan ^{2} ( x )} d x

Faktorisiere

1 - tan^{2} (x) : - (- 1 + tan^{2} (x))

= \int \frac{1}{- ( - 1 + tan ^{2} ( x ) )} d x

Entferne die Konstante:

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

= - \int \frac{1}{- 1 + tan ^{2} ( x )} d x

Wende U-Substitution an

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

Ermittle den Partialbruch von

\frac{1}{( 1 + u ^{2} ) ( - 1 + u ^{2} )} : - \frac{1}{2 ( u ^{2} + 1 )} - \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u - 1 )}

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} arctan (u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

= - (- \frac{1}{2} arctan (u) - \frac{1}{4} ln ∣ u + 1 ∣ + \frac{1}{4} ln ∣ u - 1 ∣)

Setze in

u = tan (x)

ein

= - (- \frac{1}{2} arctan (tan (x)) - \frac{1}{4} ln ∣ tan (x) + 1 ∣ + \frac{1}{4} ln ∣ tan (x) - 1 ∣)

Vereinfache

= \frac{1}{2} arctan (tan (x)) + \frac{1}{4} ln ∣ tan (x) + 1 ∣ - \frac{1}{4} ln ∣ tan (x) - 1 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{2} arctan (tan (x)) + \frac{1}{4} ln ∣ tan (x) + 1 ∣ - \frac{1}{4} ln ∣ tan (x) - 1 ∣ + C

\frac{d y}{d x} = \frac{x + y + 2}{x + y - 4}

x \to 0 lim (x^{2} + 2)

\frac{d}{d x} (cos (2 x) 2)

\int_{- 3}^{- 2} 2 x (4 - x^{2})^{3} d x

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