The method of substitution

Differentials and integrals: Integration techniques

The method of substitution

Assume that $F$ is an antiderivative of $f$ and that the function $g$ is a neat functon. The chain rule of differentiation gives:

$\frac{\dd}{\dd x}\Bigl(F\big(g(x)\bigr)\Bigr)=f\bigl(g(x)\bigr)\cdot g'(x).$ In terms of differentials:

$\dd\Bigl(F\bigl(g(x)\bigr)\Bigr)=f\bigl(g(x)\bigr)\cdot \dd\bigl(g(x)\bigr)=f\bigl(g(x)\bigr)\cdot g'(x)\,\dd x.$ This leads to the following method of substitution:

$\int f\bigl(g(x)\bigr)\, g'(x)\,\dd x=\int \dd\Bigl(F\bigl((g(x)\bigr)\Bigr)=F\bigl((g(x)\bigr)+c.$ If $u=g(x)$ and thus $\dd u=g'(x)\dd x$ , then the above rule can also be written with explicit reference to the used substitution:

Substitution Rule

$\begin{aligned}\int f\bigl(g(x)\bigr)\cdot g'(x)\,\dd x&=[u=g(x), \dd u=g'(x)\,\dd x]\\ &=\int f(u)\,\dd u\\ &=F(u)+c\\ &=F\bigl(g(x)\bigr)+c\end{aligned}$

The method of substitution is based on this rule:

For computation of $\int h(x)\,\dd x$ we look for a function $f$ and a differentiable function $g$ such that $h(x)=f\bigl(g(x)\bigr)\cdot g'(x)$ . Then we have

$\int h(x)\,\dd x=\int f(u)\,\dd u$ wuth $u=g(x)$ and is the integration of $h$ nothing else than integration of $f$ .

If $a$ and $b$ limits of integration, then:

$\int_a^b h(x)\,\dd x=\int_{g(a)}^{g(b)} f(u)\,\dd u$

This method of substitution only makes sense of course when finding an antiderivative of $f$ is an easier task than finding the antiderivative of $h$ .

A few examples illustrate the method of substitution.

Calculate the following integral:

$\int 2 t \e^ {- t^2 }\,\dd t$

We apply the substitution rule for integration with $u=t^2$ .
By differentiating $u$ we find $\dd u= 2 t \,\dd t$ .
So:

$\begin{aligned}\int 2 t \e^ {- t^2 }\,\dd t&= \int \e^ {- u }\,\dd u\\ &\phantom{uvaxyz}\blue{\text{substitution rule with }u=t^2\text{ and } \dd u= 2 t \,\dd t} \\ &= -\e^ {- u }+c\\ &\phantom{uvaxyz}\blue{\text{primitive of the integrand}}\\ &= -\e^ {- t^2 }+c\\ &\phantom{uvaxyz}\blue{\text{back substitution}}\end{aligned}$

New example

If there are limits of integration, i.e., if you want to calculate a definite integral, then you can handle the integration limits at the same time and back substitution is no longer needed. Some examples illustrate this.

Calculate the following integral via the substitution rule:

$\int_{0}^{1} {{\e^{z}}\over{\e^{2 z}+1}}\,\dd z$

We apply the substitution rule for integration with $u=\e^{z}$ .
By differentiating $u$ we find $\dd u= \e^{z} \,\dd z$ .
The integration bounds change to $e^0=1$ and $e^1=\e$ .
So

$\begin{aligned}\int_{0}^{1} {{\e^{z}}\over{\e^{2 z}+1}}\,\dd z&= \int_{1}^{\e} {{1}\over{u^2+1}}\,\dd u\\ &\phantom{uvaxyz}\blue{\text{substitution rule with }u=\e^{z}\text{ and } \dd u= \e^{z} \,\dd z} \\ &= \Bigl[\arctan \left(u\right)\Bigr]_{1}^{\e}\\ &\phantom{uvaxyz}\blue{\text{primitive of the integrand}}\\ &= \arctan \left(\e\right)-{{\pi}\over{4}} \\ &\phantom{uvaxyz}\blue{\text{substitution of integration bounds}}\end{aligned}$

New example

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Mathcentre video

Integration by Substitution (36:07)