integral from 0 to pi/3 of 1/(1-sin(x))

Lösung

\int_{0}^{\frac{π}{3}} \frac{1}{1 - sin ( x )} d x

1 + 3

Dezimale

2.73205 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{3}} \frac{1}{1 - sin ( x )} d x

Wende U-Substitution an

= \int_{0}^{\frac{3}{3}} \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}^{\frac{3}{3}} \frac{1}{1 + u ^{2} - 2 u} d u

Vervollständige das Quadrat

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

= 2 \cdot \int_{0}^{\frac{3}{3}} \frac{1}{( u - 1 ) ^{2}} d u

Wende U-Substitution an

= 2 \cdot \int_{- 1}^{\frac{1}{3} - 1} \frac{1}{v ^{2}} d v

Wende die Potenzregel an

= 2 [- \frac{1}{v}]_{- 1}^{\frac{1}{3} - 1}

Berechne die Grenzen:

\frac{3 + 3}{2} - 1

= 2 (\frac{3 + 3}{2} - 1)

Vereinfache

= 1 + 3

\int_{0}^{π} cos^{3} (θ) d θ

\int_{2}^{5} \frac{1}{( x - 2 ) ^{2}} d x

\int_{- π}^{π} sin (5 x) cos (3 x) d x

\int_{1}^{2} sin (x - 1) d x

\int_{1}^{5} 2 x ln (x^{3}) d x