高数复习（一）
高数复习
\[ \int F(x)\text{d}y=F(x)\cdot y \]
也即 \[ \int_{0}^x F(x)\text{d}y=xF(x) \]
\[ \lim\limits_{n\rightarrow\infty}\sum\limits_{i=1}^{n}\dfrac{1}{n}f(\dfrac{i}{n})=\int_{0}^{1}f(x)\text{d}x \]
积分的上下限可以通过求和上下限以及\(f\)的自变量调整.
\[ \begin{aligned} &~~~~~~ \int\dfrac{1}{2+\cos\pi x}\text{d}x\\ &=\int\dfrac{1}{2+2\cos^2\dfrac{\pi x}{2}-1}\text{d}x\\ &=\int\dfrac{1}{1+2\cos^2\dfrac{\pi x}{2}}\text{d}x\\ &=\dfrac{2}{\pi}\int\dfrac{1}{2+\sec^2\dfrac{\pi x}{2}}\text{d}(\tan\dfrac{\pi x}{2})\\ &=\dfrac{2}{\pi}\int\dfrac{1}{3+\tan^2\dfrac{\pi x}{2}}\text{d}(\tan\dfrac{\pi x}{2})\\ &= \dfrac{2}{\sqrt3}\cdot\arctan(\dfrac{\tan(\dfrac{\pi}{2}x)}{\sqrt3})+C \end{aligned} \]
\[ I=\int \frac{\cos ^2 x-\sin x}{\cos x\left(1+\cos x \cdot \mathrm{e}^{\sin x}\right)} \mathrm{d} x \]
由于 \[ \left(\cos x \cdot \mathrm{e}^{\sin x}\right)^{\prime}=-\sin x \cdot \mathrm{e}^{\sin x}+\cos ^2 x \cdot \mathrm{e}^{\sin x}=\left(\cos ^2 x-\sin x\right) \mathrm{e}^{\sin x} \] 因此 \[ \mathrm{d}\left(\cos x \cdot \mathrm{e}^{\sin x}\right)=\left(\cos ^2 x-\sin x\right) \mathrm{e}^{\sin x} \mathrm{~d} x \] 故有 \[ \begin{aligned} I & =\int \frac{\left(\cos ^2 x-\sin x\right) \mathrm{e}^{\sin x}}{\cos x \cdot \mathrm{e}^{\sin x}\left(1+\cos x \cdot \mathrm{e}^{\sin x}\right)} \mathrm{d} x=\int \frac{\mathrm{d}\left(\cos x \cdot \mathrm{e}^{\sin x}\right)}{\cos x \cdot \mathrm{e}^{\sin x}\left(1+\cos x \cdot \mathrm{e}^{\sin x}\right)} \\ & =\int\left(\frac{1}{\cos x \cdot \mathrm{e}^{\sin x}}-\frac{1}{1+\cos x \cdot \mathrm{e}^{\sin x}}\right) \mathrm{d}\left(\cos x \cdot \mathrm{e}^{\sin x}\right)=\ln \left|\frac{\cos x \cdot \mathrm{e}^{\sin x}}{1+\cos x \cdot \mathrm{e}^{\sin x}}\right|+C \end{aligned} \]