# Integration by substitution

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To recap: to integrate is to find the antiderivative of a function, i.e. to find a function that, taken the derivative of it, gives our original function back. This means that we can find integration rules just by using the derivative rules in the opposite direction.

One if the derivative rules is the chain rule:

${\left( {f(g(x))} \right)^\prime } = f'(g(x)) \cdot g'(x)$

For example

${\left( {{{({x^3} + 2)}^7}} \right)^\prime } = 7{({x^3} + 2)^6}3{x^2} = 21{({x^3} + 2)^6}{x^2}$

Working backwards that means that

$\int {21{{({x^3} + 2)}^6}{x^2}dx} = {({x^3} + 2)^7} + C$

The constant C since the derivative of an added constant is 0.

It might be hard to see how to do this, but happily there is method to do this.  This trick can be useful when the integrand (the equation we want to integrate) is a factor of two expressions.

To start with we have to look at if the derivative of a function or inner function is equal to the other factor, i.e. if we have

$\int {g(x)g'(x)} \,dx$

or

$\int {f(g(x))g'(x)} \,dx$

We call the function f (…)  the outer function and g(x) the inner function.

(The first case is just a special case of the second. The function f is just the identity function that does nothing.)

In our example above we have

$f(g(x)) = 21{({x^3} + 2)^6},\quad g'(x) = 3{x^2}$

As we will see we can be of by a constant factor, and the trick we will use will work.

The substitution

Let us do the substitution

$u = g(x)$

Now let us take the derivative of this with respect to x. We get

$u' = \frac{{du}}{{dx}} = g(x)$

Let us multiply this by dx. Can we do that? In this case we will see that it will work. The proof is in the pudding as one say.

$du = g(x)dx$

Now, this will give us

$\int {f(g(x))g'(x)} \,dx = \int {f(u)du} = F(u) + C$

Finally we replace back the expression for u:

$\int {f(g(x))g'(x)} \,dx = F(u) + C = F(g(x)) + C$

In practice

Ok, let us try this on our original example.

$\int {21{{({x^3} + 2)}^6}{x^2}dx}$

Here we can see that the derivative of x3+2 is 3x2, so we are only off with a constant factor, but that is Ok as we will see.

$\left| {\begin{array}{*{20}{c}} {u = {x^3} + 2} \\ {\frac{{du}}{{dx}} = 3{x^2}} \\ {du = 3{x^2}dx} \\ {\frac{{du}}{3} = {x^2}dx} \end{array}} \right.$

$\begin{gathered} = \int {{u^3}\frac{{du}}{4} = } \frac{1}{4}\int {{u^3}du} = \frac{1}{4}\frac{{{u^4}}}{4} + C \hfill \\ = \frac{1}{{16}}{u^4} + C = \frac{1}{{16}}{\left( {{x^4} + 5} \right)^4} + C \hfill \\ \end{gathered}$

In the above I indicate the whole process of the substitution with a vertical line, then I continue with an “=” sign, indicating that we simply continue with the integral again. In the part with the vertical line our goal is to get to something we could easily substitute away – in this case x2dx.

I would usually go directly from u=x3+2 to du=3x2dx (I would differentiate).

A few examples

$\int {{{({x^4} + 5)}^3}{x^3}dx} =$

$\left| {\begin{array}{*{20}{c}} {u = {x^4} + 5} \\ {\frac{{du}}{{dx}} = 4{x^3}} \\ {du = 4{x^3}dx} \\ {\frac{{du}}{4} = {x^3}dx} \end{array}} \right.$

$\begin{gathered} = \int {{u^3}\frac{{du}}{4} = } \frac{1}{4}\int {{u^3}du} = \frac{1}{4}\frac{{{u^4}}}{4} + C \hfill \\ = \frac{1}{{16}}{u^4} + C = \frac{1}{{16}}{\left( {{x^4} + 5} \right)^4} + C \hfill \\ \end{gathered}$

$\int {x{e^{{x^2}}}dx} = \int {{e^{{x^2}}}x{\kern 1pt} dx} =$

$\left| {\begin{array}{*{20}{c}} {u = {x^2}} \\ {\frac{{du}}{{dx}} = 2x} \\ {du = 2xdx} \\ {\frac{{du}}{2} = xdx} \end{array}} \right.$

$= \int {{e^u}\frac{{du}}{2}} = \frac{1}{2}\int {{e^u}du} = \frac{1}{2}{e^u} + C = \frac{1}{2}{e^{{x^2}}} + C$

$\int {\frac{{{x^4}}}{{{x^5} + 1}}dx} =$

Here it might be a bit harder to see what the outer function is, but it is 1 divided by something ,
i.e. the reciprocal function that has the integral ln(of the something).

$\left| {\begin{array}{*{20}{c}} {u = {x^5} + 1} \\ {\frac{{du}}{{dx}} = 5{x^4}} \\ {du = 5{x^4}dx} \\ {\frac{{du}}{5} = {x^4}dx} \end{array}} \right.$

$= \int {\frac{1}{u}\frac{{du}}{5} = \frac{1}{5}\ln (u) + C = \frac{1}{5}\ln ({x^5} + 1) + C}$

Up a level : Integrals
Previous page : Some standard integrals
Next page : Integration of definite integrals using substitutionLast modified: Jun 17, 2018 @ 19:21