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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Ermittle den Partialbruch von

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

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

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

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

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

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

Setze in

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

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d}{d x} (\frac{- x ^{2} - 9}{( x ^{2} - 9 ) ^{2}})

(x^{2} + 1) y^{^{'}} + 4 x (y - 1) = 0, y (0) = 5

\int x 16 - 25 x^{2} d x

\int_{0}^{\frac{π}{2}} 3 cos (x) d x

\int \frac{1}{( 4 x ^{2} - 25 ) ^{\frac{3}{2}}} d x