integral from 1 to 6 of-3cos((npit)/4)

解

\int_{1}^{6} - 3 cos (\frac{nπ t}{4}) d t

- \frac{3 ( 4 sin ( \frac{3 πn}{2} ) - 4 sin ( \frac{πn}{4} ) )}{πn}

解答ステップ

\int_{1}^{6} - 3 cos (\frac{nπ t}{4}) d t

定数を除く:

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

= - 3 \cdot \int_{1}^{6} cos (\frac{nπ t}{4}) d t

部分積分を適用する

= - 3 [t cos (\frac{πn t}{4}) - \int - \frac{πn}{4} t sin (\frac{πn t}{4}) d t]_{1}^{6}

\int - \frac{πn}{4} t sin (\frac{πn t}{4}) d t = t cos (\frac{πn t}{4}) - \frac{4}{πn} sin (\frac{πn t}{4})

= - 3 [t cos (\frac{πn t}{4}) - (t cos (\frac{πn t}{4}) - \frac{4}{πn} sin (\frac{πn t}{4}))]_{1}^{6}

簡素化

= - 3 [\frac{4}{πn} sin (\frac{πn t}{4})]_{1}^{6}

限度を計算する:

\frac{4}{πn} sin (\frac{3 πn}{2}) - \frac{4}{πn} sin (\frac{πn}{4})

= - 3 (\frac{4}{πn} sin (\frac{3 πn}{2}) - \frac{4}{πn} sin (\frac{πn}{4}))

簡素化

= - \frac{3 ( 4 sin ( \frac{3 πn}{2} ) - 4 sin ( \frac{πn}{4} ) )}{πn}