integral from 0 to 4 of xcos((npix)/4)

Lösung

\int_{0}^{4} x cos (\frac{nπ x}{4}) d x

\frac{16 ( - 1 ) ^{n} - 16}{π ^{2} n ^{2}}

Schritte zur Lösung

\int_{0}^{4} x cos (\frac{nπ x}{4}) d x

Wende U-Substitution an

= \int_{0}^{nπ 4} \frac{u cos ( \frac{u}{4} )}{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}^{nπ 4} u cos (\frac{u}{4}) d u

Wende die partielle Integration an

= \frac{1}{π ^{2} n ^{2}} [4 u sin (\frac{u}{4}) - \int 4 sin (\frac{u}{4}) d u]_{0}^{4 πn}

\int 4 sin (\frac{u}{4}) d u = - 16 cos (\frac{u}{4})

= \frac{1}{π ^{2} n ^{2}} [4 u sin (\frac{u}{4}) - (- 16 cos (\frac{u}{4}))]_{0}^{4 πn}

Vereinfache

= \frac{1}{π ^{2} n ^{2}} [4 u sin (\frac{u}{4}) + 16 cos (\frac{u}{4})]_{0}^{4 πn}

Berechne die Grenzen:

16 (- 1)^{n} - 16

= \frac{1}{π ^{2} n ^{2}} (16 (- 1)^{n} - 16)

Vereinfache

= \frac{16 ( - 1 ) ^{n} - 16}{π ^{2} n ^{2}}

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