integral from 0 to x of 1/(2+sin(t))

Lösung

\int_{0}^{x} \frac{1}{2 + sin ( t )} d t

\frac{2}{3} arctan (\frac{2 tan ( \frac{x}{2} ) + 1}{3}) - \frac{π}{3 3}

Schritte zur Lösung

\int_{0}^{x} \frac{1}{2 + sin ( t )} d t

Wende U-Substitution an

= \int_{0}^{t a n (\frac{x}{2})} \frac{1}{u ^{2} + 1 + u} d u

Vervollständige das Quadrat

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

= \int_{0}^{t a n (\frac{x}{2})} \frac{1}{( u + \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

Wende U-Substitution an

= \int_{\frac{1}{2}}^{t a n (\frac{x}{2}) + \frac{1}{2}} \frac{4}{4 v ^{2} + 3} d v

Entferne die Konstante:

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

= 4 \cdot \int_{\frac{1}{2}}^{t a n (\frac{x}{2}) + \frac{1}{2}} \frac{1}{4 v ^{2} + 3} d v

Wende integrale Substitution an

= 4 \cdot \int_{\frac{1}{3}}^{\frac{2}{3} (t a n (\frac{x}{2}) + \frac{1}{2})} \frac{1}{2 3 ( w ^{2} + 1 )} d w

Entferne die Konstante:

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

= 4 \cdot \frac{1}{2 3} \cdot \int_{\frac{1}{3}}^{\frac{2}{3} (t a n (\frac{x}{2}) + \frac{1}{2})} \frac{1}{w ^{2} + 1} d w

Nutze das gemeinsame Integral :

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

= 4 \cdot \frac{1}{2 3} [arctan (w)]_{\frac{1}{3}}^{\frac{2}{3} (t a n (\frac{x}{2}) + \frac{1}{2})}

Vereinfache

4 \cdot \frac{1}{2 3} [arctan (w)]_{\frac{1}{3}}^{\frac{2}{3} (t a n (\frac{x}{2}) + \frac{1}{2})} : \frac{2}{3} [arctan (w)]_{\frac{1}{3}}^{\frac{2 t a n ( \frac{x}{2} ) + 1}{3}}

= \frac{2}{3} [arctan (w)]_{\frac{1}{3}}^{\frac{2 t a n ( \frac{x}{2} ) + 1}{3}}

Berechne die Grenzen:

arctan (\frac{2 tan ( \frac{x}{2} ) + 1}{3}) - \frac{π}{6}

= \frac{2}{3} (arctan (\frac{2 tan ( \frac{x}{2} ) + 1}{3}) - \frac{π}{6})

Vereinfache

= \frac{2}{3} arctan (\frac{2 tan ( \frac{x}{2} ) + 1}{3}) - \frac{π}{3 3}

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