integral of 1/(sin(x)+cos(x))

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Vervollständige das Quadrat

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

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

Wende U-Substitution an

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 2 \cdot \frac{1}{2} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

Ersetze zurück

= 2 \cdot \frac{1}{2} \frac{ln \frac{t a n ( \frac{x}{2} ) - 1}{2} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) - 1}{2} - 1}{2}

Vereinfache

2 \cdot \frac{1}{2} \frac{ln \frac{t a n ( \frac{x}{2} ) - 1}{2} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} ) - 1}{2} - 1}{2} : 2 (\frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} + 1 - \frac{1}{2} ln \frac{tan ( \frac{x}{2} ) - 1}{2} - 1)

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

Füge eine Konstante zur Lösung hinzu

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

\int_{- 1}^{1} \frac{1}{2 x + 3} d x

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

t an g e n t f (x) = \frac{1}{x}, at x = \frac{1}{4}

\int (ln (t)) d t

t an g e n t \frac{8}{( x ^{2} + 4 )}