integral of 1/(3sin(x)+4cos(x))

Lösung

\int \frac{1}{3 sin ( x ) + 4 cos ( x )} d x

\frac{1}{5} (ln \frac{2}{5} + \frac{4}{5} tan (\frac{x}{2}) - ln - \frac{8}{5} + \frac{4}{5} tan (\frac{x}{2})) + C

Schritte zur Lösung

\int \frac{1}{3 sin ( x ) + 4 cos ( x )} d x

Wende U-Substitution an

= \int \frac{1}{3 u + 2 ( - u ^{2} + 1 )} d u

Vervollständige das Quadrat

3 u + 2 (- u^{2} + 1) : - 2 (u - \frac{3}{4})^{2} + \frac{25}{8}

= \int \frac{1}{- 2 ( u - \frac{3}{4} ) ^{2} + \frac{25}{8}} d u

Wende U-Substitution an

= \int \frac{8}{- 16 v ^{2} + 25} d v

Entferne die Konstante:

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

= 8 \cdot \int \frac{1}{- 16 v ^{2} + 25} d v

Wende integrale Substitution an

= 8 \cdot \int \frac{1}{20 ( - w ^{2} + 1 )} d w

Entferne die Konstante:

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

= 8 \cdot \frac{1}{20} \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}

= 8 \cdot \frac{1}{20} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

Ersetze zurück

= 8 \cdot \frac{1}{20} (\frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) - \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) - \frac{3}{4} ) - 1}{2})

Vereinfache

8 \cdot \frac{1}{20} (\frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) - \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) - \frac{3}{4} ) - 1}{2}) : \frac{1}{5} (ln \frac{2}{5} + \frac{4}{5} tan (\frac{x}{2}) - ln - \frac{8}{5} + \frac{4}{5} tan (\frac{x}{2}))

= \frac{1}{5} (ln \frac{2}{5} + \frac{4}{5} tan (\frac{x}{2}) - ln - \frac{8}{5} + \frac{4}{5} tan (\frac{x}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} (ln \frac{2}{5} + \frac{4}{5} tan (\frac{x}{2}) - ln - \frac{8}{5} + \frac{4}{5} tan (\frac{x}{2})) + C

\frac{d}{d x} ((5 x^{3} + 4)^{2})

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

7 t \cdot \frac{d y}{d t} + 3 y = t

\int \frac{2 e ^{2 x}}{e ^{2 x} + 14 e ^{x} + 48} d x

\frac{d y}{d x} = 3 x^{2} (1 + y^{2})