integral from-pi to pi/2 of xcos(nx)

解答

\int_{- π}^{\frac{π}{2}} x cos (n x) d x

\frac{πn sin ( \frac{πn}{2} ) + 2 cos ( \frac{πn}{2} ) - 2 ( - 1 ) ^{n}}{2 n ^{2}}

求解步骤

\int_{- π}^{\frac{π}{2}} x cos (n x) d x

使用换元积分法

= \int_{- πn}^{\frac{πn}{2}} \frac{u cos ( u )}{n ^{2}} d u

提出常数:

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

= \frac{1}{n ^{2}} \cdot \int_{- πn}^{\frac{πn}{2}} u cos (u) d u

使用分布积分法

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

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

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

化简

= \frac{1}{n ^{2}} [u sin (u) + cos (u)]_{- πn}^{\frac{πn}{2}}

带入上下限计算后得:

\frac{πn}{2} sin (\frac{πn}{2}) + cos (\frac{πn}{2}) - (- 1)^{n}

= \frac{1}{n ^{2}} (\frac{πn}{2} sin (\frac{πn}{2}) + cos (\frac{πn}{2}) - (- 1)^{n})

化简

= \frac{πn sin ( \frac{πn}{2} ) + 2 cos ( \frac{πn}{2} ) - 2 ( - 1 ) ^{n}}{2 n ^{2}}

\int_{- 2}^{0} 12 x^{3} - 14 x^{2} d x

\int_{- 2}^{2} x + 1 d x

\int_{- 1}^{2} e^{x + 1} d x

\int_{0}^{1} \frac{x}{1 + 6 x ^{2}} d x

\int_{0}^{4} (x^{2} + x) d x