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受欢迎的微积分 >

integral from 1 to 2 of xcos((npix)/2)

解答

\int_{1}^{2} x cos (\frac{nπ x}{2}) d x

解答

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

求解步骤

\int_{1}^{2} x cos (\frac{nπ x}{2}) d x

使用换元积分法

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

提出常数:

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

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

使用换元积分法

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{\frac{πn}{2}}^{πn} 4 v cos (v) d v

提出常数:

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

= \frac{1}{π ^{2} n ^{2}} \cdot 4 \cdot \int_{\frac{πn}{2}}^{πn} v cos (v) d v

使用分布积分法

= \frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - \int sin (v) d v]_{\frac{πn}{2}}^{πn}

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

= \frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{\frac{πn}{2}}^{πn}

化简

\frac{1}{π ^{2} n ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{\frac{πn}{2}}^{πn} : \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{\frac{πn}{2}}^{πn}

= \frac{4}{π ^{2} n ^{2}} [v sin (v) + cos (v)]_{\frac{πn}{2}}^{πn}

带入上下限计算后得:

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

= \frac{4}{π ^{2} n ^{2}} ((- 1)^{n} - \frac{πn}{2} sin (\frac{πn}{2}) - cos (\frac{πn}{2}))

化简

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

流行的例子

integral from 0 to pi/6 of (cos(3θ))^2

\int_{0}^{\frac{π}{6}} (cos (3 θ))^{2} d θ

integral from 1 to 5 of (1-x)(x-5)

\int_{1}^{5} (1 - x) (x - 5) d x

integral from 0 to y of x/y

\int_{0}^{y} \frac{x}{y} d x

integral from 1 to 2 of (10-3x)/(x^3)

\int_{1}^{2} \frac{10 - 3 x}{x ^{3}} d x

integral from 0 to 1 of pi(1-x)^2

\int_{0}^{1} π (1 - x)^{2} d x