\begin{equation} \int_0^{\sqrt{3} a} \frac{d x}{d^2+x^2} \end{equation}
时间: 2023-03-24 11:01:31 浏览: 52
首先,我们需要解出二阶导数$\frac{d^2x}{dx^2}$:
\begin{align*}
\frac{d}{dx}\left(\frac{dx}{dt}\right) &= \frac{d^2x}{dt^2}\\
\frac{d}{dx}\left(\frac{dx}{dt}\right) &= \frac{d}{dx}\left(\sqrt{3}a\sin\left(\frac{2\pi}{T}t\right)\right)\\
\frac{d^2x}{dt^2} &= \frac{d}{dt}\left(\frac{d}{dt}\sqrt{3}a\sin\left(\frac{2\pi}{T}t\right)\right)\\
&= \frac{d}{dt}\left(\sqrt{3}a\cdot\frac{2\pi}{T}\cos\left(\frac{2\pi}{T}t\right)\right)\\
&= -\frac{4\pi^2}{T^2}\sqrt{3}a\sin\left(\frac{2\pi}{T}t\right)
\end{align*}
现在我们可以计算出定积分:
\begin{align*}
\int_0^T\sqrt{3}a\frac{dx}{d^2x/dt^2}dt &= \int_0^T\sqrt{3}a\frac{1}{d^2x/dt^2}d\left(\frac{dx}{dt}\right)\\
&= \int_0^T\sqrt{3}a\frac{1}{d^2x/dt^2}\frac{d^2x}{dt^2}dt\\
&= \int_0^T\sqrt{3}a\cdot\frac{-1}{4\pi^2/T^2}\sin\left(\frac{2\pi}{T}t\right)dt\\
&= \frac{-\sqrt{3}aT}{4\pi}\int_0^1\sin(2\pi u)du\qquad\text{(令}u=\frac{t}{T}\text{)}\\
&= \frac{-\sqrt{3}aT}{4\pi}\left[-\frac{1}{2\pi}\cos(2\pi u)\right]_0^1\\
&= \frac{\sqrt{3}aT}{8}
\end{align*}
因此,答案为$\frac{\sqrt{3}aT}{8}$。