题目 03
$$
I=\int_{0}^{\pi} x \sqrt{\cos ^{2} x-\cos ^{4} x} \mathrm{~ d} x=?
$$
解析 03
已知有如下公式:
$$
\int_{0}^{\pi} x f(\sin x) \mathrm{~ d} x=\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \mathrm{~ d} x
$$
于是:
$$
I=\int_{0}^{\pi} x \sqrt{\cos ^{2}-\cos ^{4} x} \mathrm{~ d} x=\int_{0}^{\pi} x \sqrt{\cos ^{2} x\left(1-\cos ^{2} x\right)} \mathrm{~ d} x =
$$
$$
\int_{0}^{\pi} x \sqrt{\left(1-\sin ^{2} x\right) \sin ^{2} x} \mathrm{~ d} x=\frac{\pi}{2} \int_{0}^{\pi} \sqrt{\left(1-\sin ^{2} x\right) \sin ^{2} x} \mathrm{~ d} x \Rightarrow
$$
又:
$$
x \in(0, \pi) \Rightarrow \sin x>0 \Rightarrow
$$
$$
I=\frac{\pi}{2} \int_{0}^{\pi} \sin x \sqrt{1-\sin ^{2} x} \mathrm{~ d} x
$$
在上面的计算中极易发生计算错误,主要原因是根号带有一定的绝对值效果,因此,去根号的时候,一定要注意验证正负性(下面的示例步骤是错误):
$$
I=\frac{\pi}{2} \int_{0}^{\pi} \sin x \sqrt{\cos ^{2} x} \mathrm{~ d} x=\frac{\pi}{2} \int_{0}^{\pi} \sin x \cos x \mathrm{~ d} x =
$$
$$
\frac{\pi}{2} \int_{0}^{\pi} \sin x \mathrm{~ d} (\sin x)=\frac{\pi}{2} \times\left.\frac{1}{2} \sin ^{2} x\right|_{0} ^{\pi}=
$$
$$
\frac{\pi}{4}(0-0)=\frac{\pi}{4} \times 0=0
$$
正确的求解步骤为:
$$
I=\frac{\pi}{2} \int_{0}^{\pi} \sin x \sqrt{\cos ^{2} x} \mathrm{~ d} x \Rightarrow
$$
$$
x \in\left(0, \frac{\pi}{2}\right) \Rightarrow \cos x>0, \quad x \in\left(\frac{\pi}{2}, \pi\right) \Rightarrow \cos x<0 \Rightarrow
$$
$$
I=\frac{\pi}{2} \int_{0}^{\pi} \sin x|\cos x| \mathrm{~ d} x \Rightarrow
$$
$$
I=\frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} \sin x \cos x \mathrm{~ d} x-\frac{\pi}{2} \int_{\frac{\pi}{2}}^{\pi} \sin x \cos x \mathrm{~ d} x \Rightarrow
$$
$$
I=\frac{\pi}{2}\left[\int_{0}^{\frac{\pi}{2}} \sin x \mathrm{~ d} (\sin x)-\int_{\frac{\pi}{2}}^{\pi} \sin x \mathrm{~ d} (\sin x)\right] \Rightarrow
$$
$$
I=\frac{\pi}{2}\left[\left.\frac{1}{2} \sin ^{2} x\right|_{0} ^{\frac{\pi}{2}}-\left.\frac{1}{2} \sin ^{2} x\right|_{\frac{\pi}{2}} ^{\pi}\right] \Rightarrow
$$
$$
I=\frac{\pi}{2}\left[\frac{1}{2}(1-0)-\frac{1}{2}(0-1)\right] \Rightarrow
$$
$$
I=\frac{\pi}{2}\left(\frac{1}{2}+\frac{1}{2}\right)=\frac{\pi}{2}
$$