integral from 0 to 1 of tcos(npit)

Lösung

\int_{0}^{1} t cos (nπ t) d t

\frac{( - 1 ) ^{n} - 1}{π ^{2} n ^{2}}

Schritte zur Lösung

\int_{0}^{1} t cos (nπ t) d t

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die partielle Integration an

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

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

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

Vereinfache

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

Berechne die Grenzen:

(- 1)^{n} - 1

= \frac{1}{π ^{2} n ^{2}} ((- 1)^{n} - 1)

Vereinfache

= \frac{( - 1 ) ^{n} - 1}{π ^{2} n ^{2}}

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