带有三角函数的积分不容易计算怎么办？尝试把三角函数放到微分符号 d 里面，这样就可以用整体代换法去掉三角函数了

一、题目

$$
I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+x) \cos x}{1+\cos ^{2} x} \mathrm{~d} x = ?
$$

难度评级：

二、解析

$$
I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+x) \cos x}{1+\cos ^{2} x} \mathrm{~ d} x \Rightarrow
$$

$$
I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x+x \cos x}{1+\cos ^{2} x} \mathrm{~ d} x \Rightarrow
$$

$$
I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+\cos ^{2} x} \mathrm{~ d} x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x \cos x}{1+\cos ^{2} x} \mathrm{~ d} x.
$$

由于 $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x \cos x}{1+\cos ^{2} x} \mathrm{~ d} x$ 是奇函数，因此：

$$
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x \cos x}{1+\cos ^{2} x} \mathrm{~ d} x = 0
$$

即：

$$
I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+\cos ^{2} x} \mathrm{~ d} x.
$$

又：

$$
1+\cos ^{2} x+\sin ^{2} x-\sin ^{2} x=2-\sin ^{2} x
$$

$$
\cos x = \mathrm{d} \left(\sin x\right)
$$

即：

$$
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+\cos ^{2} x} \mathrm{~ d} x=
$$

$$
\int_{-1}^{1} \frac{\mathrm{d} \left( \sin x \right)}{2-\sin ^{2} x}=
$$

令 $x = \sin x$, 则：

$$
\int_{-1}^{1} \frac{1}{2-x^{2}} \mathrm{~ d} x=2 \int_{0}^{1} \frac{1}{(\sqrt{2}+x)(\sqrt{2}-x)} \mathrm{~ d} x
$$

又：

$$
\frac{1}{(\sqrt{2}+x)(\sqrt{2}-x)}=\left[\frac{1}{\sqrt{2}+x}+\frac{1}{\sqrt{2}-x}\right] \cdot \frac{1}{2 \sqrt{2}}
$$

于是：

$$
2 \int_{0}^{1} \frac{1}{(\sqrt{2}+x)(\sqrt{2}-x)} \mathrm{~ d} x=
$$

$$
\frac{1}{\sqrt{2}} \int_{0}^{1}\left[\frac{1}{\sqrt{2}+x}+\frac{1}{\sqrt{2}-x}\right] \mathrm{~ d} x
$$

$$
\left.\frac{1}{\sqrt{2}}[\ln (\sqrt{2}+x)-\ln (\sqrt{2}-x)]\right|_{0} ^{1}=
$$

$$
\frac{1}{\sqrt{2}}[\ln (\sqrt{2}+1)-\ln (\sqrt{2}-1)-\ln (\sqrt{2})+\ln (\sqrt{2})] =
$$

$$
\frac{1}{\sqrt{2}}[\ln (\sqrt{2}+1)-\ln (\sqrt{2}-1)]=
$$

$$
\frac{1}{\sqrt{2}} \ln \frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} \ln \frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=
$$

$$
\frac{\sqrt{2}}{2} \ln \frac{(\sqrt{2}+1)^{2}}{1}=
$$

$$
\frac{\sqrt{2}}{2} \ln (\sqrt{2}+1)^{2}=\sqrt{2} \ln (\sqrt{2}+1).
$$

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