integral from 1 to e^pi of sin(ln(x))

解答

\int_{1}^{e^{π}} sin (ln (x)) d x

\frac{e ^{π} + 1}{2}

十进制

12.07034 \dots

求解步骤

\int_{1}^{e^{π}} sin (ln (x)) d x

使用分布积分法

= [x sin (ln (x)) - \int cos (ln (x)) d x]_{1}^{e^{π}}

\int cos (ln (x)) d x = \frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x))

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

化简

[x sin (ln (x)) - (\frac{1}{2} x sin (ln (x)) + \frac{1}{2} x cos (ln (x)))]_{1}^{e^{π}} : [\frac{1}{2} (x sin (ln (x)) - x cos (ln (x)))]_{1}^{e^{π}}

= [\frac{1}{2} (x sin (ln (x)) - x cos (ln (x)))]_{1}^{e^{π}}

带入上下限计算后得:

\frac{e ^{π}}{2} + \frac{1}{2}

= \frac{e ^{π}}{2} + \frac{1}{2}

化简

= \frac{e ^{π} + 1}{2}

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