2.积分

1 \(\int cscx cot x dx = -csc x + C\)

记忆时：为什么又负号，首先\(cscx = \frac {1}{sin x}\) 分母为正 \((cscx)' = \frac {-cosx}{(sinx)^2} = - cscx cotx\)

2.\((arcsinx)'\)

• y = arcsinx
• 反函数: x = siny
• 公式：函数的导数 = \(\frac {1}{(反函数))'}\)
• = \(y' = \frac{1}{(siny)'} = \frac {1}{cosy} = \frac {1}{\sqrt{1-sin^2y}},(x = siny) y' = \frac{1}{\sqrt{1-x^2}}\)
• 这里y = arctanx, y = arccosx都是一样推导

\(\int tanx^2x dx\)

= tanx - x + c
\((tanx - x)'\)
= \(\frac {sin^2x + cos^2x}{cos^2x} - 1\)

\(\int sectdt\)

• =\(|\ln|secx + tanx|+C\)

关于分数什么时候积分为ln

首先\(\int frac {1}{x} = lnx + c\)对于这种分母为简单的一次项

三角公式

\(1 + tan^2x = sex^2x\)

• 推导
  • = \(1 + \frac {sin^2}{cos^2}\)
  • = \(\frac {cos^2 x + sin^2x}{cos^2x}\)
  • = \(\frac {1}{cos^2x}\)
  • = \(sec^2x\)

\(\tan x 与\sin x的转化\)

令\(\tan \frac {x}{2} = t\)

\(sin x = 2 * sin\frac{x}{2} * cos\frac{x}{2}\)，分母化成1

\(\frac {2 * sin\frac{x}{2} * cos\frac{x}{2}}{1} = \frac {2 * sin\frac{x}{2} * cos\frac{x}{2}}{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}}\),同除以\(sin\frac{x}{2} * cos\frac{x}{2}\)

=\(\frac {2}{\tan \frac{x}{2}+\cot \frac{x}{2}} = \frac{2}{t + \frac{1}{t}}\)分子分母同乘以t

= \(\frac {2t}{t^2 + 1}\)

tanx与cosx

同上，end = \(\frac {1 - t^2}{1 + t^2}\)