integral of sqrt(1+csc(x))

Lösung

\int 1 + csc (x) d x

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

Schritte zur Lösung

\int 1 + csc (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int 1 + \frac{1}{sin ( x )} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

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

Wende die Summenregel an:

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

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

\int \frac{u}{u ( 1 + u ^{2} )} d u = 22 (- \frac{1}{8} (ln 2 u + 22 u + 2 - 2 arctan (2 u + 1)) + \frac{1}{8} (ln 2 u - 22 u + 2 + 2 arctan (2 u - 1)))

\int \frac{1}{u ( 1 + u ^{2} )} d u = 2 (\frac{1}{4 2} (ln 2 u + 22 u + 2 + 2 arctan (2 u + 1)) + \frac{1}{4 2} (- ln 2 u - 22 u + 2 + 2 arctan (2 u - 1)))

= 2 (22 (- \frac{1}{8} (ln 2 u + 22 u + 2 - 2 arctan (2 u + 1)) + \frac{1}{8} (ln 2 u - 22 u + 2 + 2 arctan (2 u - 1))) + 2 (\frac{1}{4 2} (ln 2 u + 22 u + 2 + 2 arctan (2 u + 1)) + \frac{1}{4 2} (- ln 2 u - 22 u + 2 + 2 arctan (2 u - 1))))

Setze in

u = tan (\frac{x}{2})

ein

= 2 (22 (- \frac{1}{8} (ln 2 tan (\frac{x}{2}) + 22 tan (\frac{x}{2}) + 2 - 2 arctan (2 tan (\frac{x}{2}) + 1)) + \frac{1}{8} (ln 2 tan (\frac{x}{2}) - 22 tan (\frac{x}{2}) + 2 + 2 arctan (2 tan (\frac{x}{2}) - 1))) + 2 (\frac{1}{4 2} (ln 2 tan (\frac{x}{2}) + 22 tan (\frac{x}{2}) + 2 + 2 arctan (2 tan (\frac{x}{2}) + 1)) + \frac{1}{4 2} (- ln 2 tan (\frac{x}{2}) - 22 tan (\frac{x}{2}) + 2 + 2 arctan (2 tan (\frac{x}{2}) - 1))))

Multipliziere aus

22 (- \frac{1}{8} (ln 2 tan (\frac{x}{2}) + 22 tan (\frac{x}{2}) + 2 - 2 arctan (2 tan (\frac{x}{2}) + 1)) + \frac{1}{8} (ln 2 tan (\frac{x}{2}) - 22 tan (\frac{x}{2}) + 2 + 2 arctan (2 tan (\frac{x}{2}) - 1))) + 2 (\frac{1}{4 2} (ln 2 tan (\frac{x}{2}) + 22 tan (\frac{x}{2}) + 2 + 2 arctan (2 tan (\frac{x}{2}) + 1)) + \frac{1}{4 2} (- ln 2 tan (\frac{x}{2}) - 22 tan (\frac{x}{2}) + 2 + 2 arctan (2 tan (\frac{x}{2}) - 1))) : 2 arctan (2 tan (\frac{x}{2}) + 1) + 2 arctan (2 tan (\frac{x}{2}) - 1)

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

Füge eine Konstante zur Lösung hinzu

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

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

\int \frac{x ^{3} - 1}{x} d x

\int \frac{2 y}{- 2 x y + x ^{2}} d y

\int \frac{12 x ^{2} - 4 x}{x ^{2}} d x

\int x^{3} cos (3 x^{4} + 2) d x