integral from-pi to pi of x^2cos(kx)

解答

\int_{- π}^{π} x^{2} cos (k x) d x

\frac{4 π ( - 1 ) ^{k}}{k ^{2}}

求解步骤

\int_{- π}^{π} x^{2} cos (k x) d x

使用换元积分法

= \int_{- πk}^{kπ} \frac{u ^{2} cos ( u )}{k ^{3}} d u

提出常数:

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

= \frac{1}{k ^{3}} \cdot \int_{- πk}^{kπ} u^{2} cos (u) d u

使用分布积分法

= \frac{1}{k ^{3}} [u^{2} sin (u) - \int 2 u sin (u) d u]_{- πk}^{kπ}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{k ^{3}} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{- πk}^{kπ}

带入上下限计算后得:

4 π (- 1)^{k} k

= \frac{1}{k ^{3}} \cdot 4 π (- 1)^{k} k

化简

= \frac{4 π ( - 1 ) ^{k}}{k ^{2}}

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