Computing integrals (2): Integration by substitution

 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.  $\int f(x)dx=\int f(x(t))x^{\prime }(t)dt$. 
2. For any $a,b\in J$, ${\int }_{x(a)}^{x(b)}f(x)dx={\int }_a^{b}f(x(t))x^{\prime }(t)dt$.

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

$$\frac{d}{dt}F(x(t))=F^{\prime }(x(t))x^{\prime }(t)=f(x(t))x^{\prime }(t).$$

Thus $F(x(t))$ as a function of $t$ is an anti-derivative of $f(x(t))x^{\prime }(t)$. Thus Part 1 is proven. Part 2 follows immediately. ■

Example. Let $f(x)$ be a differentiable function. Then

$$\int \frac{f^{\prime }(x)}{f(x)}dx=\log |f(x)|+C.$$

Proving this is an exercise. □

Example. Let us compute the anti-derivative of $\int \frac{dx}{(ax+b{)}^m}$ where $a\ne 0$ and $m\in \mathbb{N}$.

Let $z=ax+b$. Then $\frac{dx}{dz}=1/a$ so $dx=\frac{dz}{a}$.

$$\int \frac{dx}{(ax+b{)}^m}=\frac{1}{a}\int \frac{dz}{z^m}=\{\begin{array}{cc}\frac{1}{a(1-m)z^{m-1}}+C& (m\ge 2),\\ \frac{1}{a}\log |z|+C& (m=1).\end{array}$$

Thus,

$$\int \frac{dx}{(ax+b{)}^m}=\{\begin{array}{cc}\frac{1}{a(1-m)(ax+b{)}^{m-1}}+C& (m\ge 2),\\ \frac{1}{a}\log |ax+b|+C& (m=1).\end{array}$$

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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,

$$\begin{array}{rcl}\int x\sqrt{1-x}dx& =& \int (1-t^2)t(-2t)dt\\ & =& 2\int (t^4-t^2)dt\\ & =& 2(\frac{t^5}{5}-\frac{t^3}{3})+C\\ & =& \frac{2}{15}{t}^3(3t^2-5)+C\\ & =& \frac{2}{15}(1-x{)}^{\frac{3}{2}}[3(1-x)-5]+C\\ & =& -\frac{2}{15}(1-x{)}^{\frac{3}{2}}(3x+2)+C.\end{array}$$

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