integral from-pi to 0 of xcos((2n+1)x)

Lösung

\int_{- π}^{0} x cos ((2 n + 1) x) d x

\frac{1 - cos ( π ( 1 + 2 n ) ) - π sin ( π ( 1 + 2 n ) ) - 2 πn sin ( π ( 1 + 2 n ) )}{( 2 n + 1 ) ^{2}}

Schritte zur Lösung

\int_{- π}^{0} x cos ((2 n + 1) x) d x

Wende U-Substitution an

= \int_{- π (2 n + 1)}^{0} \frac{u cos ( u )}{( 2 n + 1 ) ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{( 2 n + 1 ) ^{2}} \cdot \int_{- π (2 n + 1)}^{0} u cos (u) d u

Wende die partielle Integration an

= \frac{1}{( 2 n + 1 ) ^{2}} [u sin (u) - \int sin (u) d u]_{- π (2 n + 1)}^{0}

\int sin (u) d u = - cos (u)

= \frac{1}{( 2 n + 1 ) ^{2}} [u sin (u) - (- cos (u))]_{- π (2 n + 1)}^{0}

Vereinfache

= \frac{1}{( 2 n + 1 ) ^{2}} [u sin (u) + cos (u)]_{- π (2 n + 1)}^{0}

Berechne die Grenzen:

1 - cos (π (1 + 2 n)) - π sin (π (1 + 2 n)) - 2 πn sin (π (1 + 2 n))

= \frac{1}{( 2 n + 1 ) ^{2}} (1 - cos (π (1 + 2 n)) - π sin (π (1 + 2 n)) - 2 πn sin (π (1 + 2 n)))

Vereinfache

= \frac{1 - cos ( π ( 1 + 2 n ) ) - π sin ( π ( 1 + 2 n ) ) - 2 πn sin ( π ( 1 + 2 n ) )}{( 2 n + 1 ) ^{2}}

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