integral from 0 to 1.5 of 1/(1+sin(x))

Lösung

\int_{0}^{1.5} \frac{1}{1 + sin ( x )} d x

0.96458 \dots

Schritte zur Lösung

\int_{0}^{1.5} \frac{1}{1 + sin ( x )} d x

Wende U-Substitution an

= \int_{0}^{0.93159 \dots} \frac{2}{1 + u ^{2} + 2 u} d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{0.93159 \dots} \frac{1}{1 + u ^{2} + 2 u} d u

Vervollständige das Quadrat

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

= 2 \cdot \int_{0}^{0.93159 \dots} \frac{1}{( u + 1 ) ^{2}} d u

Wende U-Substitution an

= 2 \cdot \int_{1}^{1.93159 \dots} \frac{1}{v ^{2}} d v

Wende die Potenzregel an

= 2 [- \frac{1}{v}]_{1}^{1.93159 \dots}

Berechne die Grenzen:

0.48229 \dots

= 2 \cdot 0.48229 \dots

Vereinfache

= 0.96458 \dots

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