substitution for integrationSubstitutionForIntegration2004-08-27 05:24:282009-07-01 04:23:31Theoremantiderivative% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}For determining the antiderivative $F(x)$ of a given real function $f(x)$ in a ``closed form'', i.e. for integrating $f(x)$, the result is often obtained by using the
\begin{thmplain}
\,\,If
$$\int f(x)\,dx = F(x)+C$$
and \,$x = x(t)$\, is a differentiable function,
then
\begin{align}
F(x(t)) = \int f(x(t))\,x'(t)\,dt+c.
\end{align}
\end{thmplain}
{\em Proof.} \, By virtue of the chain rule,
$$\frac{d}{dt}F(x(t)) = F'(x(t))\cdot x'(t),$$
and according to the supposition, $F'(x) = f(x)$. \,Thus we get the claimed equation (1).\\
\textbf{Remarks.}
\begin{itemize}
\item The expression $x'(t)\,dt$ in (1) may be understood as the differential of $x(t)$.
\item For returning to the original variable $x$, the inverse function \,$t = t(x)$\, of $x(t)$ must be substituted to $F(x(t))$.\\
\end{itemize}
\textbf{Example.} \, For integrating $\int \frac{x\,dx}{1+x^4}$ we take \,$x^2 = t$\, as a new variable. \,Then, \,$2x\,dx = dt$, $x\,dx = \frac{dt}{2}$, and we get
$$\int \frac{x\,dx}{1+x^4} = \frac{1}{2}\int \frac{dt}{1+t^2} = \frac{1}{2}\arctan t+ C= \frac{1}{2}\arctan x^2+C.$$