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

Lösung

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

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

Schritte zur Lösung

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

Vereinfache

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

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

= - \frac{1}{2 2} \cdot \int - \frac{1}{v ^{2} + 1} + 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

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

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

\int 1 d v = v

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

Ersetze zurück

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

Vereinfache

- \frac{1}{2 2} (- arctan (2 cot (x)) + 2 cot (x)) : \frac{1}{2 2} arctan (2 cot (x)) - \frac{1}{2} cot (x)

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

Füge eine Konstante zur Lösung hinzu

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

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

\int x \cdot cos (a \cdot x) d x

\int 3 t^{2} (t^{3} + 2)^{\frac{3}{2}} d t

\int \frac{17 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

\int \frac{( x ^{2} + 3 x - 2 )}{( x - 2 ) ( x ^{2} + 7 )} d x