\documentclass[12pt]{article}
\title{\hei �L�n��}
\author{\kai �L���u}
\date{1 April 1999}
\newtheorem{exmp}{��}

\begin{document}

\centerline{\hei ��\hbox to 2truecm{\hfill}�D} 
�p��U�C�U�n���G
\begin{tabbing}
\hbox to 6cm{\hfill}\=\kill
1. $\int \cos^7 x\sin x \,dx$. \>
2. $\int e^x \sin x\,dx$. \\
3. $\int_1^2 x^3 \ln x\,dx$. \>
4. $\int_1^2 \tan^{-1} x\,dx$. \\ 
5. $\int \cos^{-1} x\,dx$. \>
6. $\int x^2 \cos x\,dx$. \\
\end{tabbing}

\section{\hei �N��.}

�]��� $u$ �b $[a,b]$ �W�㦳�@�����s��ɨ��,�Y $f$ �O�ӳs����,�B $f$ ���w�q
��]�t $u$ ���Ȱ�,�O $F$ �� $f$ ���Ͼɨ��,�h�I�ηL���Ǫ��s��ߩ� $F(u(x))$ �o
\[ \frac{d}{dx}F(u(x)) = F'(u(x))u'(x) = f(u(x))u'(x). \]
�q�Ӧ�
\[ \int f(u(x))u'(x)\,dx = F(u(x)) + c = \int f(u)\,du, \hbox to 1truecm{\hfill}u=u(x). \]

�o�N�O���w�n�����N������.�۷����w�n�����N�������h��
\[ \int_a^b f(u(x))u'(x)\,dx = \int_{u(a)}^{u(b)}f(u)\,du. \]\\

���|�X�Ӫ����N����²��Ҥl:
\begin{exmp}
$\int \sin^3 x \cos x \, dx$.
\end{exmp}\\

��. �] $u(x)=sin{x}$,�h$u'(x)=cos{x}$. \\

�즡 =$ \int (u(x))^3 \cdot u'(x)\,dx = \int u^3\,du = \frac{u^4}{4} + c = \frac{\sin^4 x}{4} + c$. 
\begin{exmp}
�D$\int x \sqrt{a^2 + x^2}\,dx$.
\end{exmp}\\

��. �] $u(x) = a^2 + x^2$,�h $u'(x) = 2x$.\\

�즡 = $ \frac{1}{2} \int u(x)^{1/2} \cdot u'(x)\,dx = \frac{1}{2} \int u^{1/2}\,du = \frac{1}{2} \cdot \frac{2}{3} \cdot u^{3/2} + c = \frac{1}{3}(a^2 + x^2)^{3/2} + c$.
\begin{exmp}
$\int \frac{dx}{x(\ln x)^2}$.
\end{exmp}\\

��. �O $u = \ln x$.�h�즡 = $-1/u + c = -\frac{1}{\ln x} + c$.
\begin{exmp}
�D $\int \tan x\,dx$.
\end{exmp}\\

��. �] $u(x) = \cos x$,�h
\[ \int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = - \int \frac{u'(x)}{u(x)}\,dx = - \int \frac{du}{u} = -\ln|u| + c = \ln|\sec x| + c. \]
�A�|�@�ǤޤJ�s�ܼƪ��Ҥl.
\begin{exmp}
�D $\int_0^1 \frac{du}{(1+u^2)^2}$.
\end{exmp}\\

��. �] $u = \tan x$,$0\leq x\leq\pi/2$,�h$du/dx = \sec^2 x$.�D��
\[ \int_0^1 \frac{du}{(1+u^2)^2} = \int_{\arctan0}^{\arctan1} \frac{\sec^2 x}{(1 + \tan^2 x)^2}\,dx = \int_0^{\pi/4} \frac{dx}{\sec^2 x} \]
\[\hbox to 2truecm{\hfill} = \frac{1}{2} \int_0^{\pi/4} 1 + \cos 2x\,dx = (\frac{x}{2} + \frac{\sin 2x}{4})|_0^{\frac{\pi}{4}} \]
\[ = \frac{\pi}{8} + \frac{1}{4}. \hbox to 3truecm{\hfill} \]\\
Ū�̽бN���D�P�W�`����11���.
\begin{exmp}
�D $\int_1^2 \frac{dx}{x \sqrt{1+x}}$.
\end{exmp}\\

��. �] $\sqrt{1+x} = y$,�Y$x = y^2-1$,�h$dx/dy = 2y$.��O
\[ \int \frac{dx}{x \sqrt{1+x}} = \int \frac{2y}{(y^2-1) \cdot y}\,dy = \int (\frac{1}{y-1}-\frac{1}{y+1})\,dy \]
\[ \hbox to 1.5truecm{\hfill}= \ln |\frac{y-1}{y+1}| + c = \ln |\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}| + c. \]\\
�G
\[ \int_1^2 \frac{dx}{x\sqrt{1+x}} = \ln \frac{\sqrt{3}-1}{\sqrt{3}+1} - \ln \frac{\sqrt{2}-1}{\sqrt{2}+1}. \]\\

�b�W�����p��L�{��,�ڭ̧Q�Ψ�@�ӧޥ�
\[ \frac{2}{y^2-1} = \frac{1}{y-1} - \frac {1}{y+1}. \]

�o�K�O�U�@�`��������������.

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