integral from 0 to pi of x^4cos(nx)

解

\int_{0}^{π} x^{4} cos (n x) d x

- \frac{4 ( 6 π ( - 1 ) ^{n} - π ^{3} ( - 1 ) ^{n} n ^{2} )}{n ^{4}}

解答ステップ

\int_{0}^{π} x^{4} cos (n x) d x

部分積分を適用する

= [\frac{1}{n} (x^{4} sin (n x) - n \cdot \int \frac{4}{n} x^{3} sin (n x) d x)]_{0}^{π}

\int \frac{4}{n} x^{3} sin (n x) d x = \frac{4}{n} (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x))))

= [\frac{1}{n} (x^{4} sin (n x) - n \frac{4}{n} (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x)))))]_{0}^{π}

簡素化

[\frac{1}{n} (x^{4} sin (n x) - n \frac{4}{n} (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x)))))]_{0}^{π} : [\frac{1}{n} (x^{4} sin (n x) - 4 (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x)))))]_{0}^{π}

= [\frac{1}{n} (x^{4} sin (n x) - 4 (- \frac{1}{n} x^{3} cos (n x) + \frac{3}{n ^{4}} (n^{2} x^{2} sin (n x) - 2 (- n x cos (n x) + sin (n x)))))]_{0}^{π}

限度を計算する:

- \frac{4}{n} (\frac{6 π ( - 1 ) ^{n}}{n ^{3}} - (- 1)^{n} \frac{π ^{3}}{n})

= - \frac{4}{n} (\frac{6 π ( - 1 ) ^{n}}{n ^{3}} - (- 1)^{n} \frac{π ^{3}}{n})

簡素化

= - \frac{4 ( 6 π ( - 1 ) ^{n} - π ^{3} ( - 1 ) ^{n} n ^{2} )}{n ^{4}}