integral from 0 to 1 of 2xy^2sec^2(3xy)

Lösung

\int_{0}^{1} 2 x y^{2} sec^{2} (3 x y) d x

\frac{2 y tan ( 3 y )}{3} + \frac{2 ln ∣ cos ( 3 y ) ∣}{9}

Schritte zur Lösung

\int_{0}^{1} 2 x y^{2} sec^{2} (3 x y) d x

Entferne die Konstante:

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

= 2 y^{2} \cdot \int_{0}^{1} sec^{2} (3 y x) x d x

Wende die partielle Integration an

= 2 y^{2} [\frac{1}{3 y} x tan (3 y x) - \int \frac{1}{3 y} tan (3 y x) d x]_{0}^{1}

\int \frac{1}{3 y} tan (3 y x) d x = - \frac{1}{9 y ^{2}} ln ∣ cos (3 y x) ∣

= 2 y^{2} [\frac{1}{3 y} x tan (3 y x) - (- \frac{1}{9 y ^{2}} ln ∣ cos (3 y x) ∣)]_{0}^{1}

Vereinfache

= 2 y^{2} [\frac{1}{3 y} x tan (3 y x) + \frac{1}{9 y ^{2}} ln ∣ cos (3 y x) ∣]_{0}^{1}

Berechne die Grenzen:

\frac{1}{3 y} tan (3 y) + \frac{1}{9 y ^{2}} ln ∣ cos (3 y) ∣

= 2 y^{2} (\frac{1}{3 y} tan (3 y) + \frac{1}{9 y ^{2}} ln ∣ cos (3 y) ∣)

Vereinfache

= \frac{2 y tan ( 3 y )}{3} + \frac{2 ln ∣ cos ( 3 y ) ∣}{9}

\int_{0}^{2} ln (3 - x) d x

\int_{- 1}^{1} x arccos (x) d x

\int_{0}^{2} x^{2} - 40 d x

\int_{1}^{3} x^{2} - 3 x + 1 d x

\int_{3}^{4} (3 x^{2} - 1) d x