«求定积分 $\int ^1_{-1}\sqrt{\frac{1+x}{1-x}}\rm{d}x$ (对称平均值积分)»
by pluvet on Dec 14, 2019

$$ \begin{align} f(x) &= \sqrt{\frac{1+x}{1-x}}\\ f(-x)&= \sqrt{\frac{1-x}{1+x}}\\ f(x)+f(-x) &= \sqrt{\frac{1+x}{1-x}} + \sqrt{\frac{1-x}{1+x}}\\ &=\frac{2}{\sqrt{1-x^2}}\\ \int^{1}_{-1}f(x){\rm{d}}x &= \frac{1}{2}\int^{1}_{-1}(f(x)+f(-x)){\rm{d}}x\\ &=\int^{1}_{-1}\frac{1}{\sqrt{1-x^2}}{\rm{d}}x\\ &=\arcsin(1) - \arcsin(-1)\\& = \pi \end{align} $$

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