Computing integrals (2): Integration by substitution

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 By using the chain rule, we may compute the integral of complicated (composite) functions.



Theorem (Integration by substitution)

Let \(f(x)\) be a continuous function. Suppose \(x\) is a differentiable function of \(t\) on an open interval \(J\), \(x = x(t)\). Then

  1.  \(\int f(x)dx = \int f(x(t))x'(t)dt\). 
  2. For any \(a,b\in J\), \(\int_{x(a)}^{x(b)}f(x)dx = \int_a^bf(x(t))x'(t)dt\).
Remark. In Part 1, ``\(=\)'' means that the left-hand side is equal to the right-hand side except for a constant term. □
Proof. Let us fix \(a \in J\). For any \(t \in J\), let us define the anti-derivative
\[F(x(t)) = \int_{x(a)}^{x(t)}f(x)dx.\]
Up to this point, we consider \(F(x(t))\) as a function of \(x(t)\) (the independent variable is \(x(t)\), and \(t\) is just a parameter). 

Now regard \(F(x(t))\) as a function of \(t\) and differentiate it with respect to \(t\) using the chain rule. We have
\[\frac{d}{dt}F(x(t)) = F'(x(t))x'(t) = f(x(t))x'(t). \]
Thus \(F(x(t))\) as a function of \(t\) is an anti-derivative of \(f(x(t))x'(t)\). Thus Part 1 is proven. Part 2 follows immediately. ■

Example. Let \(f(x)\) be a differentiable function. Then
\[\int\frac{f'(x)}{f(x)}dx = \log|f(x)| + C.\]
Proving this is an exercise. □

Example. Let us compute the anti-derivative of \(\int \frac{dx}{(ax + b)^m}\) where \(a \neq 0\) and \(m\in \mathbb{N}\).
Let \(z = ax + b\). Then \(\frac{dx}{dz} = 1/a\) so \(dx = \frac{dz}{a}\).
\[ \int \frac{dx}{(ax+b)^m} = \frac{1}{a}\int\frac{dz}{z^m} = \left\{ \begin{array}{cc} \frac{1}{a(1-m)z^{m-1}} + C & (m \geq 2),\\ \frac{1}{a}\log|z| + C & (m = 1). \end{array}\right.\]
Thus,
\[ \int \frac{dx}{(ax+b)^m} =\left\{ \begin{array}{cc} \frac{1}{a(1-m)(ax+b)^{m-1}} + C & (m \geq 2),\\ \frac{1}{a}\log|ax+b| + C & (m = 1). \end{array}\right. \]
□

Example. Let us compute \(\int x\sqrt{1 -x}dx\).
Let \(t = \sqrt{1-x}\). Then \(t^2 = 1 -x\) so \(x\sqrt{1 - x} = (1-t^2)t\). Then \(\frac{d}{dt}t^2 = - \frac{dx}{dt}\) so \(dx = (-2t)dt\). Thus,
\[ \begin{eqnarray*} \int x\sqrt{1-x}dx &=& \int(1-t^2)t(-2t)dt\\ &=& 2\int(t^4 - t^2)dt\\ &=& 2\left(\frac{t^5}{5} - \frac{t^3}{3}\right) + C\\ &=& \frac{2}{15}t^3(3t^2 - 5) + C\\ &=& \frac{2}{15}(1-x)^{\frac{3}{2}}[3(1-x) - 5] + C\\ &=& -\frac{2}{15}(1-x)^{\frac{3}{2}}(3x + 2) + C. \end{eqnarray*} \]
□

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