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

Lösung

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

2 arctan (\frac{2 tan ( x )}{1 - tan ^{2} ( x )}) - arcsin (tan (x)) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Trigonometrische Substitution anwenden

= \int \frac{cos ^{2} ( v )}{1 + sin ^{2} ( v )} d v

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

\frac{1 - sin ^{2} ( v )}{1 + sin ^{2} ( v )} = \frac{2}{1 + sin ^{2} ( v )} - 1

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

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{2}{1 + sin ^{2} ( v )} d v - \int 1 d v

\int \frac{2}{1 + sin ^{2} ( v )} d v = 2 arctan (2 tan (v))

\int 1 d v = v

= 2 arctan (2 tan (v)) - v

Ersetze zurück

= 2 arctan (2 tan (arcsin (tan (x)))) - arcsin (tan (x))

Vereinfache

2 arctan (2 tan (arcsin (tan (x)))) - arcsin (tan (x)) : 2 arctan (\frac{2 tan ( x )}{1 - tan ^{2} ( x )}) - arcsin (tan (x))

= 2 arctan (\frac{2 tan ( x )}{1 - tan ^{2} ( x )}) - arcsin (tan (x))

Füge eine Konstante zur Lösung hinzu

= 2 arctan (\frac{2 tan ( x )}{1 - tan ^{2} ( x )}) - arcsin (tan (x)) + C

d er i v a t i v e (x^{3} + 4) e^{x}

t an g e n t f (x) = (1 - x) (x^{2} - 3) 2

x \to - 1 lim (x^{3})

\frac{\partial}{\partial x} (7 y)

\int_{1}^{5} (6 - x) - (8 - 3 x) d x