integral from 1 to e^{pi/2 of}sin(ln(θ))

解答

\int_{1}^{e^{\frac{π}{2}}} sin (ln (θ)) d θ

\frac{e ^{\frac{π}{2}} + 1}{2}

十进制

2.90523 \dots

求解步骤

\int_{1}^{e^{\frac{π}{2}}} sin (ln (θ)) d θ

使用分布积分法

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

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

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

化简

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

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

带入上下限计算后得:

\frac{e ^{\frac{π}{2}}}{2} + \frac{1}{2}

= \frac{e ^{\frac{π}{2}}}{2} + \frac{1}{2}

化简

= \frac{e ^{\frac{π}{2}} + 1}{2}

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