integral from 0 to pi of 2xsin(nx)

解答

\int_{0}^{π} 2 x sin (n x) d x

- \frac{2 π ( - 1 ) ^{n}}{n}

求解步骤

\int_{0}^{π} 2 x sin (n x) d x

提出常数:

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

= 2 \cdot \int_{0}^{π} x sin (n x) d x

使用分布积分法

= 2 [\frac{1}{n} (- x cos (n x) - n \cdot \int - \frac{1}{n} cos (n x) d x)]_{0}^{π}

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

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

化简

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

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

带入上下限计算后得:

- (- 1)^{n} \frac{π}{n}

= 2 (- (- 1)^{n} \frac{π}{n})

化简

= - \frac{2 π ( - 1 ) ^{n}}{n}

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