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Mathematical Expression Editor
We explore more difficult problems involving substitution.
We begin by restating the substitution formula.
Integral Substitution Formula If \(g\) is differentiable on the interval \([a,b]\) and \(f\) is
differentiable on the interval \([g(a),g(b)]\), and if they have continuous derivatives, then
\[ \int _a^b f'(g(x)) g'(x) \d x =\int _{g(a)}^{g(b)} f'(u) \d u. \]
Sometimes it is not obvious how a fraction could have been obtained using the chain
rule. A common trick though is to substitute for the denominator of a fraction. Like
all tricks, this technique does not always work. Regardless the next two examples
present how this technique can be used.
when \(\sec (x) \neq - \tan (x)\). So in a very contrived way, we have just proved
\[ \int \sec (x) \d x = \ln (|\sec (x) + \tan (x)|) + C. \]
Notice the variable in this next example.
Compute:
\[ \int \frac {u}{1-u^2} \d u \]
We want to substitute for \(1-u^2\). But the variable “\(u\)” has already been
used…OH NO! Never fear! We can substitute with whatever variable that
we want. In particular, let us use “\(w\)” for this problem. So we let
\[ w = 1 - u^2 \]
and then
\begin{align*} \d w &= \answer [given]{-2u} \d u,\\ \d u &= \answer [given]{\frac {-1}{2u}} \d w. \end{align*}
Thus
\begin{align*} \int \frac {u}{1-u^2} \d u &= \int \frac {u}{w} \left ( \answer [given]{\frac {-1}{2u}} \right ) \d w \\ &= - \frac {1}{2} \int \frac {1}{w} \d w \\ &= - \frac {1}{2} \ln (|w|) + C \\ &= - \frac {1}{2} \ln (|1-u^2|) + C. \end{align*}
Now, if we make the substitution \(u = e^x\), we have that
\begin{align*} \d u &= \answer [given]{e^x} \d x,\\ \d x &= \frac {1}{\answer [given]{e^x}} \d u, \end{align*}
and
\begin{align*} \int \frac {e^{2x}}{1 - e^{2x}} \d x &= \int \frac {(e^x)^2}{1 - u^2} \cdot \frac {1}{\answer [given]{e^x}} \d u \\ &= \int \frac {e^x}{1-u^2} \d u \\ &= \int \frac {u}{1-u^2} \d u. \end{align*}
But now we are back to Example key example, and so we know that
\begin{align*} \int \frac {e^{2x}}{1 - e^{2x}} \d x &= - \frac {1}{2} \ln (|1-u^2|) + C \\ &= - \frac {1}{2} \ln (|1 - e^{2x}|) + C. \end{align*}
Again, in the previous example we could have instead made the substitution
\[ u = 1 - e^{2x} \]
and
avoided using Example key example. In general, any time that you make two successive
substitutions in a problem, you could have instead just made one substitution. This
one substitution is the composition of the two original substitutions. But
sometimes it may not be obvious to make one clever substitution, and so two
substitutions makes more sense. The next example helps to demonstrate
this.
Compute:
\[ \int _0^{16} \sqrt {4 - \sqrt {x}} \d x \]
While it is not obvious at all, let us try the substitution
\[ u = \sqrt {x}. \]
Then
\begin{align*} \d u &= \answer [given]{\frac {1}{2 \sqrt {x}}} \d x,\\ \d x &= 2 \sqrt {x} \d u = 2u \d u, \end{align*}