4-解题技巧.tex

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\begin{document}
\begin{enumerate}
  \item $\frac{\sin 2t }{1+ \cos 2t}=\tan t$ \\
  \item $1 \pm \sin x ={(\sin \frac{x}{2} \pm \cos \frac{x}{2})}^2$ \\
  \item $\int \frac{1}{1+ \sin t }dt = \int \frac{1-\sin t}{1-{\sin t}^2}=\int \sec^2 tdt$ \\
  \item $\int 1 dx =\int \sqrt{2x+1} d \sqrt{2x+1}$ \\
  \item $\int \ln{(1+ \tan t)} dt = \int \ln \frac{\cos t +\sin t}{\cos t}dt=
  \int \ln \frac{\cos t + \cos (\frac{\pi}{2} -t ) }{\cos t} dt =
  \int \ln \frac{2 \cos \frac{\pi}{4} \cos (\frac{\pi}{4} -t)}{\cos t}dt =
$和差化积 \\
  \item $\int \frac{1}{\sin^3 t}dt =
\int \frac{\sin^2 t + \cos^2 t}{\sin^3 t }dt$
  \item $\int \cot^2 \frac{x}{2} dx= \int \csc^2 \frac{x}{2} -1 dx $
  \item $ \int \frac{1}{1+\sin^2 x} dx =\int \frac{\frac{1}{\cos^2 x}}{\frac {1}{\cos^2 x } + \tan^2 t}
  = \int \frac {d \tan t}{1+ 2\tan^2 x}
  = \frac{1}{\sqrt{2}} \int \frac{d ( \sqrt {2} \tan x)}{1+ {(\sqrt{2} \tan x)}^2}$
  \item $\int \frac{1}{1+t^3}dt=\int_0^x\frac{1-t^2+t^2}{1+t^3}dt$
  \item $\int \frac{1}{1+t^3}dt=\frac{1}{3} \int\frac{1}{1+x} -\frac{1}{6}\int \frac{2x-1}{x^2-x+1}dx +\frac{1}{2} \int \frac{1}{x^2-x+1} dx$
  \item $\int \frac{dx}{{(1+x^2)}^\frac{3}{2}}=\frac{x}{\sqrt{1+x^2}} +C $ \\
  \item 方程 $x^2 -Xx+Y=0 ;X=x_1+x_2 ; Y=x_1 \dot x_2 $


\end{enumerate}
\end{document}