integral of ((sin(x))/(1+tan(x)))^2

Lösung

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

- \frac{1}{4 ( 1 + tan ( x ) )} - \frac{1}{4} cos^{2} (x) - \frac{1}{4} arctan (tan (x)) + C

Schritte zur Lösung

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

Trigonometrische Substitution anwenden

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

Ermittle den Partialbruch von

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

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

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

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

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

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

Setze in

u = tan (x)

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int csc^{2} (2 a x) d x

\int x e^{- inπ x} d x

\int \frac{1}{1 + 8 x} d x

\int sec^{4} (a) x d x

\int \frac{x + 7}{x ^{2} + 2 x} d x