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对称性在定积分上的使用

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分类:

数学技巧

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考研 高等数学 定积分 对称性
目录

问题
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Vy=2πa3∫02π(t−sin⁡t)(1−cos⁡t)(1−cos⁡t)dt V_y = 2 \pi a^3 \int_{0}^{2 \pi} (t - \sin t)(1 - \cos t)(1 - \cos t) \mathop{}\mathrm{d} t

想法
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显然，展开后求定积分即可。群友们也是这样做的。

但思来想去，总感觉应该有更快的方法，然后发现了这么个做法：

Vy=2πa3∫02π(t−sin⁡t)(1−cos⁡t)(1−cos⁡t)dt=8πa3∫02π(t−π−sin⁡t+π)cos⁡4t2dt=8πa3[∫02π(t−π)cos⁡4t2dt−∫02πsin⁡tcos⁡4t2dt+∫02ππcos⁡4t2dt] \begin{aligned} V_y &= 2 \pi a^3 \int_{0}^{2 \pi} (t - \sin t)(1 - \cos t)(1 - \cos t) \mathop{}\mathrm{d} t \\ &= 8 \pi a^3 \int_{0}^{2 \pi } (t - \pi - \sin t + \pi) \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t \\ &= 8 \pi a^3 \left[ \int_{0}^{2 \pi } (t - \pi) \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t - \int_{0}^{2 \pi } \sin t \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t + \int_{0}^{2 \pi } \pi \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t \right] \end{aligned} Vy=8π2a3∫02πcos⁡4t2dt \begin{aligned} V_y &= 8 \pi^2 a^3 \int_{0}^{2 \pi } \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t \end{aligned} Vy=16π2a3∫0πcos⁡4t2dt=32π2a3∫0π2cos⁡4udu=32π2a3⋅34⋅12⋅π2=6π3a3 \begin{aligned} V_y &= 16 \pi^2 a^3 \int_{0}^{ \pi } \cos^4 \frac{t}{2} \mathop{}\mathrm{d} t \\ &= 32 \pi^2 a^3 \int_{0}^{ \frac{\pi}{2} } \cos^4 u \mathop{}\mathrm{d} u \\ &= 32 \pi^2 a^3 \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \\ &= 6 \pi^3a^3 \end{aligned}

启发
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∫0πxf(sin⁡x)dx=π2∫0πf(sin⁡x)dxf(x) 连续 \int_{0}^{\pi} xf(\sin x) \mathop{}\mathrm{d} x = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \mathop{}\mathrm{d} x \qquad f(x)\text{ 连续}

从直观上来证明的话，就是因为 sin⁡x\sin x 是关于 π2\displaystyle \frac{\pi}{2} 对称的，而且 f(x)f(x) 是连续的，所以 f(sin⁡x)f(\sin x) 也是关于 π2\displaystyle \frac{\pi}{2} 对称的。而 x−π2x - \displaystyle \frac{\pi}{2} 则关于 (π2,0)\left(\displaystyle \frac{\pi}{2} , 0\right) 对称

g(x)=(x−π2)f(sin⁡x) g(x) = (x - \frac{\pi}{2}) f(\sin x) g(π2+x)=xf(cos⁡x)=−[−xf(cos⁡x)]=−g(π2−x) g(\frac{\pi}{2} + x) = x f(\cos x) = -[-xf(\cos x)] = -g(\frac{\pi}{2} - x) ∫0πxf(sin⁡x)dx=∫0π(x−π2+π2)f(sin⁡x)dx=π2∫0πf(sin⁡x)dx \int_{0}^{\pi} xf(\sin x) \mathop{}\mathrm{d} x = \int_{0}^{\pi} (x - \frac{\pi}{2} + \frac{\pi}{2})f(\sin x) \mathop{}\mathrm{d} x = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \mathop{}\mathrm{d} x