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

Lösung

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

\frac{π}{3 3}

Dezimale

0.60459 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende U-Substitution an

= \int_{\frac{1}{2}}^{\frac{3}{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}}^{\frac{3}{2}} \frac{1}{4 v ^{2} + 3} d v

Wende integrale Substitution an

= 4 \cdot \int_{\frac{1}{3}}^{3} \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}}^{3} \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}}^{3}

Vereinfache

4 \cdot \frac{1}{2 3} [arctan (w)]_{\frac{1}{3}}^{3} : \frac{2}{3} [arctan (w)]_{\frac{1}{3}}^{3}

= \frac{2}{3} [arctan (w)]_{\frac{1}{3}}^{3}

Berechne die Grenzen:

\frac{π}{6}

= \frac{2}{3} \cdot \frac{π}{6}

Vereinfache

= \frac{π}{3 3}

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