kaoyan3basic 高等数学 第73题
📝 题目
### 第73题 $\displaystyle 73 I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+x) \cos x}{1+\cos ^{2} x} \mathrm{~d} x=$ $\_\_\_\_$ ．
💡 答案解析
**答案**：$\displaystyle \frac{\pi}{2}$ **解析**： 步骤1：$\displaystyle I = \int_{-\pi/2}^{\pi/2} \frac{(1+x)\cos x}{1+\cos^2 x} \mathrm{d}x = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+\cos^2 x} \mathrm{d}x + \int_{-\pi/2}^{\pi/2} \frac{x\cos x}{1+\cos^2 x} \mathrm{d}x$。 步骤2：第二项被积函数为奇函数，在对称区间积分为0。第一项被积函数为偶函数，故$\displaystyle I = 2\int_{0}^{\pi/2} \frac{\cos x}{1+\cos^2 x} \mathrm{d}x$。 步骤3：令$t = \sin x$，则$\mathrm{d}t = \cos x \mathrm{d}x$，$x=0$时$t=0$，$x=\pi/2$时$t=1$，$1+\cos^2 x = 1+1-\sin^2 x = 2 - t^2$，原式$\displaystyle = 2\int_{0}^{1} \frac{1}{2-t^2} \mathrm{d}t = 2 \cdot \frac{1}{2\sqrt{2}} \ln\left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|\Big|_{0}^{1} = \frac{1}{\sqrt{2}} \ln\frac{\sqrt{2}+1}{\sqrt{2}-1} = \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1)^2 = \sqrt{2}\ln(\sqrt{2}+1)$。 **答案更正**：$\displaystyle \int_{0}^{1} \frac{1}{2-t^2} \mathrm{d}t = \frac{1}{2\sqrt{2}}\ln\frac{\sqrt{2}+t}{\sqrt{2}-t}\Big|_{0}^{1} = \frac{1}{2\sqrt{2}}\ln\frac{\sqrt{2}+1}{\sqrt{2}-1}$，乘以2得$\displaystyle \frac{1}{\sqrt{2}}\ln\frac{\sqrt{2}+1}{\sqrt{2}-1} = \frac{1}{\sqrt{2}}\ln(\sqrt{2}+1)^2 = \sqrt{2}\ln(\sqrt{2}+1)$。 **最终答案**：$\sqrt{2}\ln(\sqrt{2}+1)$ **难度**：★★★☆☆