题目 05
已知:
$$
f(x)=x-\int_{0}^{\pi} f(x) \cos x \mathrm{~ d} x
$$
则:
$$
I=\int_{0}^{\pi} f(x) \sin ^{4} x \mathrm{~ d} x=?
$$
解析 05
解法一
令:
$$
A=\int_{0}^{\pi} f(x) \cos x \mathrm{~ d} x \Rightarrow
$$
$$
f(x)=x-A \Rightarrow
$$
$$
I=\int_{0}^{\pi} f(x) \sin ^{4} x \mathrm{~ d} x=\int_{0}^{\pi}(x-A) \sin ^{4} \mathrm{~ d} x=
$$
$$
\int_{0}^{\pi} x \sin ^{4} \mathrm{~ d} x-A \int_{0}^{\pi} \sin ^{4} \mathrm{~ d} x=
$$
$$
\frac{\pi}{2} \times 2 \times \int_{0}^{\frac{\pi}{2}} \sin ^{4} \mathrm{~ d} x-A \times 2 \times \int_{0}^{\frac{\pi}{2}} \sin ^{4} \mathrm{~ d} x=
$$
$$
\frac{\pi}{2} \times 2 \times\left(\frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right)-2 A \times\left(\frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right) \Rightarrow
$$
$$
I=(\pi-2 A) \times \frac{3 \pi}{16}
$$
于是,把“未知”往“已知”的形式上凑:
$$
f(x)=x-\int_{0}^{\pi} f(x) \cos x \mathrm{~ d} x \Rightarrow
$$
$$
f(x) \cos x=x \cos x-\cos x \int_{0}^{\pi} f(x) \cos x \mathrm{~ d} x \Rightarrow
$$
两边同时积分:
$$
\int_{0}^{\pi} f(x) \cos x=\int_{0}^{\pi} x \cos x \mathrm{~ d} x\left[\int_{0}^{\pi} f(x) \cos x \mathrm{~ d} x\right] \cdot \int_{0}^{\pi} \cos x \mathrm{~ d} x
$$
$$
A=\int_{0}^{\pi} x \cos x \mathrm{~ d} x-A \int_{0}^{\pi} \cos x \mathrm{~ d} x
$$
又:
$$
\int_{0}^{\pi} \cos x \mathrm{~ d} x=\left.\sin x\right|_{0} ^{\pi}=(0-0)=0
$$
$$
\int_{0}^{\pi} x \cos x \mathrm{~ d} x=\int_{0}^{\pi} x \mathrm{~ d} (\sin x)=
$$
$$
\left.x \sin x\right|_{0} ^{\pi}-\int_{0}^{\pi} \sin x \mathrm{~ d} x=-\int_{0}^{\pi} \sin x \mathrm{~ d} x=
$$
$$
-\left[-\left.\cos x\right|_{0} ^{\pi}\right]=\cos \pi-\cos 0=-1-1=-2 \Rightarrow
$$
于是:
$$
A=-2-A \times 0 \Rightarrow A=-2 \Rightarrow
$$
$$
I=(\pi+4) \times \frac{3 \pi}{16} .
$$
解法二
$$
f(x)=x-\int_{0}^{\pi} f(x) \cos x \mathrm{~ d} x \Rightarrow
$$
$$
f(x)=x-\int_{0}^{\pi} f(x) \mathrm{~ d} (\sin x) \Rightarrow
$$
$$
f(x)=x-\left[\left.f(x) \cdot \sin x\right|_{0} ^{\pi}-\int_{0}^{\pi} \sin x f^{\prime}(x) \mathrm{~ d} x\right] \Rightarrow
$$
$$
f(x)=x-\left[0-\int_{0}^{\pi} \sin x \mathrm{~ d} x\right] \Rightarrow
$$
$$
f(x)=x+\left(-\left.\cos x\right|_{0} ^{\pi}\right) \Rightarrow
$$
$$
f(x)=x+[-(-1-1)] \Rightarrow f(x)=x+2 \Rightarrow
$$
于是:
$$
I=\int_{0}^{\pi} f(x) \sin ^{4} x \mathrm{~ d} x=\int_{0}^{\pi}(x+2) \sin ^{4} x \mathrm{~ d} x=
$$
$$
\int_{0}^{\pi} x \sin ^{4} x \mathrm{~ d} x+2 \int_{0}^{\pi} \sin ^{4} x \mathrm{~ d} x \Rightarrow
$$
$$
I=\frac{\pi}{2} \times 2 \times \int_{0}^{\frac{\pi}{2}} \sin ^{4} x \mathrm{~ d} x+2 \times 2 \int_{0}^{\frac{\pi}{2}} \sin ^{4} x \mathrm{~ d} x \Rightarrow
$$
$$
I=\pi \times\left(\frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right)+4 \times\left(\frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right) \Rightarrow
$$
$$
I=(\pi+4) \times \frac{3 \pi}{16}
$$