三角函数积分公式¶
一¶
\[ \int_{0}^{\tfrac{\pi}{2}} f(\sin x) \mathrm{d}x = \int_{0}^{\tfrac{\pi}{2}} f(\cos x) \mathrm{d}x \]
用 区间再现公式，即可证明。
二¶
\[ \int_{0}^{\pi} xf(\sin x) \mathrm{d}x = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \mathrm{d}x \]
用 区间再现公式 > 推论，即可证明。
三¶
\[ \begin{align} \int_{0}^{\pi} f(\sin x) \mathrm{d}x &= 2\int_{0}^{\tfrac{\pi}{2}} f(\sin x) \mathrm{d}x \\ \\ \int_{0}^{\pi} xf(\sin x) \mathrm{d}x &= \pi \int_{0}^{\tfrac{\pi}{2}} f(\sin x) \mathrm{d}x \end{align} \]
对称性。
四¶
\[ \int_{0}^{\tfrac{\pi}{2}} x(f(\sin x) + f(\cos x)) \mathrm{d}x = \frac{\pi}{2} \int_{0}^{\tfrac{\pi}{2}} f(\sin x) \mathrm{d}x \]
证明¶
\[ \begin{align} \int_{0}^{\tfrac{\pi}{2}} xf(\sin x) \mathrm{d}x &= \int_{0}^{\tfrac{\pi}{2}} (\frac{\pi}{2} - x)f(\sin (\frac{\pi}{2} - x)) \mathrm{d}x \\ \\ &= \frac{\pi}{2} \int_{0}^{\tfrac{\pi}{2}} f(\cos x) \mathrm{d}x - \int_{0}^{\tfrac{\pi}{2}} xf(\cos x) \mathrm{d}x \end{align} \]
等式两边再加上
\[ \int_{0}^{\tfrac{\pi}{2}} xf(\cos x) \mathrm{d}x \]
就得到结论了。
五¶
\[ \begin{align} \int \sec x \mathrm{d}x &= \int \dfrac{\sec x(\sec x + \tan x)}{\sec x + \tan x} \mathrm{d}x \\ \\ &= \int \dfrac{\mathrm{d} (\sec x + \tan x)}{\sec x + \tan x} \\ \\ &= \ln \left | \sec x + \tan x \right | + C \end{align} \]
或者
\[ \begin{align} \int \sec x \mathrm{d}x &= \int \dfrac{\cos x}{\cos^2 x} \mathrm{d}x \\ \\ &= \int \dfrac{\mathrm{d} \sin x}{1 - \sin^2 x} \\ \\ &= \dfrac{1}{2} \int (\dfrac{1}{1 + \sin x} + \dfrac{1}{1 - \sin x}) \mathrm{d} \sin x \\ \\ &= \dfrac{1}{2} \ln \left | \dfrac{1 + \sin x}{1 - \sin x} \right | + C \end{align} \]
六¶
\[ \begin{align} \int \csc x \mathrm{d}x &= \int \dfrac{1}{\sin x} \mathrm{d}x \\ \\ &= \int \dfrac{1}{2 \sin \dfrac{x}{2} \cos \dfrac{x}{2}} \mathrm{d}x \\ \\ &= \int \dfrac{\cos \dfrac{x}{2}}{\sin \dfrac{x}{2} \cos^2 \dfrac{x}{2}} \mathrm{d} \dfrac{x}{2} \\ \\ &= \int \dfrac{\sec^2 \dfrac{x}{2}}{\tan \dfrac{x}{2}} \mathrm{d} \dfrac{x}{2} \\ \\ &= \int \dfrac{1}{\tan \dfrac{x}{2}} \mathrm{d} \tan \dfrac{x}{2} \\ \\ &= \ln \left | \tan \dfrac{x}{2} \right | + C \end{align} \]
由于
\[ \begin{align} \tan \dfrac{x}{2} &= \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \\ \\ &= \dfrac{2 \sin^2 \dfrac{x}{2}}{2 \sin \dfrac{x}{2} \cos \dfrac{x}{2}} \\ \\ &= \dfrac{1 - \cos x}{\sin x} \\ \\ &= \csc x - \cot x \end{align} \]
所以，也有
\[ \int \csc x \mathrm{d}x = \ln \left | \csc x - \cot x \right | + C \]