integral from-pi to 0 of (pi+x)cos(nx)

解答

\int_{- π}^{0} (π + x) cos (n x) d x

\frac{1 - ( - 1 ) ^{n}}{n ^{2}}

求解步骤

\int_{- π}^{0} (π + x) cos (n x) d x

使用分布积分法

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

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

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

化简

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

带入上下限计算后得:

\frac{1}{n ^{2}} - (- 1)^{n} \frac{1}{n ^{2}}

= \frac{1}{n ^{2}} - (- 1)^{n} \frac{1}{n ^{2}}

化简

= \frac{1 - ( - 1 ) ^{n}}{n ^{2}}

\int_{0}^{zy} (z) d z

\int_{0}^{1} 50 x (2 x^{2} + 1)^{4} d x

\int_{1}^{13} ln (t) d t

\int_{0}^{3} x^{5} d x

\int_{1}^{3} \frac{2 x}{( x ^{2} + 3 ) ^{2}} d x