integral from 0 to pi of xcos^2(x)

解答

\int_{0}^{π} x cos^{2} (x) d x

\frac{π ^{2}}{4}

十进制

2.46740 \dots

求解步骤

\int_{0}^{π} x cos^{2} (x) d x

使用分布积分法

= [\frac{1}{4} (x (2 x + sin (2 x)) - 4 \cdot \int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x)]_{0}^{π}

\int \frac{1}{2} (x + \frac{1}{2} sin (2 x)) d x = \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x))

= [\frac{1}{4} (x (2 x + sin (2 x)) - 4 \cdot \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x)))]_{0}^{π}

化简

[\frac{1}{4} (x (2 x + sin (2 x)) - 4 \cdot \frac{1}{2} (\frac{x ^{2}}{2} - \frac{1}{4} cos (2 x)))]_{0}^{π} : [\frac{1}{8} (2 x^{2} + 2 x sin (2 x) + cos (2 x))]_{0}^{π}

= [\frac{1}{8} (2 x^{2} + 2 x sin (2 x) + cos (2 x))]_{0}^{π}

带入上下限计算后得:

\frac{2 π ^{2} + 1}{8} - \frac{1}{8}

= \frac{2 π ^{2} + 1}{8} - \frac{1}{8}

化简

= \frac{π ^{2}}{4}

x \to 0 lim ((1 - 7 x)^{\frac{1}{x}})

\int \frac{( x ^{3} - 4 x )}{( x ^{2} + 1 ) ^{2}} d x

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x \to \infty + lim (\frac{1}{x})

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