计算极限 $\underset{n\to \infty }{\lim }\frac{\sqrt[n]{2024}\left(1-\cos \frac{1}{n^2}\right)n^3}{\sqrt{1+n^2}-n}$

$$\begin{array}{rl}\underset{n\to \infty }{\lim }\frac{\sqrt[n]{2024}\left(1-\cos \frac{1}{n^2}\right)n^3}{\sqrt{1+n^2}-n}& =\underset{n\to \infty }{\lim }\frac{\frac{1}{2}{\left(\frac{1}{n^2}\right)}^2{n}^3\left(\sqrt{1+n^2}+n\right)}{1}\underset{n\to \infty }{\lim }\sqrt[n]{2024}\\ & =1\end{array}$$

计算定积分 ${\int }_0^{\pi }{\cos }^{2}x{e}^x\mathrm{d}x$

$$\begin{array}{rl}{\int }_0^{\pi }{\cos }^{2}x{e}^x\mathrm{d}x+{\int }_0^{\pi }{\sin }^{2}x{e}^x\mathrm{d}x& ={\int }_0^{\pi }{e}^x\mathrm{d}x\\ & =e^{\pi }-1\\ {\int }_0^{\pi }{\cos }^{2}x{e}^x\mathrm{d}x-{\int }_0^{\pi }{\sin }^{2}x{e}^x\mathrm{d}x& ={\int }_0^{\pi }\cos 2x{e}^x\mathrm{d}x\\ & =\left(k_1\cos 2x+k_2\sin 2x\right)e^x{|}_0^{\pi }\end{array}$$

由方程组 $k_1+2k_2=1$，$k_2-2k_1=0$ 可得 $k_1=\frac{1}{5}$，$k_2=\frac{2}{5}$。

$$\begin{array}{rl}\left(\frac{1}{5}\cos 2x+\frac{2}{5}\sin 2x\right)e^x{|}_0^{\pi }& =\frac{1}{5}\left(e^{\pi }-1\right)\end{array}$$

故原式为 $\frac{3}{5}\left(e^{\pi }-1\right)$。

计算曲线积分
$${\oint }_C\frac{y}{\sqrt{1+x^2}}\mathrm{d}x+\left(4x+\ln \left(x+\sqrt{1+x^2}\right)\right)\mathrm{d}y$$

其中曲线 $C$ 为从 $A(1,0)$ 到 $B(-1,0)$ 的上半圆周，方向为逆时针。

$$\begin{array}{rl}{\oint }_C\frac{y}{\sqrt{1+x^2}}\mathrm{d}x+\ln \left(x+\sqrt{1+x^2}\right)\mathrm{d}y& ={\oint }_{C}4x\mathrm{d}y\\ & ={\int }_0^{\pi }4{\cos }^{}\\ & \\ & \end{array}$$