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Integration by Substitution

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It is fairly rare that there is a function which we can integrate directly, as for a non-basic function it is hard to reverse engineer a derivative in our heads. This means that we need to use an integration method. One method is integration by substitution.

Integration by substitution is where we introduce a new variable into the equation, and we use this choice of substitution to make the integral easier to solve.

Whenever we make a substitution, we must also change the differential (let's call it \(\mathrm{d}x\)) in the integral, as we will now be considering the area with respect to a change in a different variable. We do this by differentiating the substitution, and then treating the differential of the substitution (eg \( \mathrm{d}x / \mathrm{d}y \)) like a fraction, to replace the original differential.

A good way to view integration by substitution is the inverse of the chain rule for differentiation. As a reminder, the chain rule for two functions f, g is given as \( \left[f (g (x))\right] '= f' (g (x)) \cdot g '(x)\).

To get to the basic formula for integration by parts, we can integrate both sides with respect to \(x\) to get

\[ \int f'(g(x))g'(x) \, \mathrm{d}x = f(g(x)).\]

We then let \(u = g (x)\) and then we will get to the integral of \(f '(u)\), which is simply \(f (u) = f (g (x))\).

If we can integrate by substitution it is usually beneficial to do so, rather than resort to another method such as integration by parts. This is because substitution is often a quicker and more efficient method than integration by parts. However most integrals cannot be solved both ways, so it is essential to have a good knowledge of both.

The general method for performing integration by substitution is as follows:

Choose a substitution that will allow all terms to be changed and make the integral as easy as possible.

Differentiate the substitution so that we can change the differential.

Perform the substitution.

Complete the integral.

Undo the substitution.

Use integration by substitution to integrate

\[ \int \frac{2x+7}{x^2+7x+14} \, \mathrm{d}x.\]

Let us try to first try a substitution of \(u = x^2 + 7x+14\). If we use this substitution, then

\[ \frac{\mathrm{d} u}{\mathrm{d}x} = 2x+7 \]

so

\[ \mathrm{d}x = \frac{\mathrm{d} u }{2x+7}.\]

Now let's substitute this in

\[ \begin{align} \int \frac{2x+7}{x^2+7x+14} \, \mathrm{d}x &= \int \frac{2x+7}{u}\frac{\mathrm{d} u }{2x+7} \\ &= \int\frac{1}{u}\, \mathrm{d} u .\end{align} \]

This is now a standard integral, so we can integrate this. As we have already seen, this integrates to \( \ln | u | \).

\[ \begin{align} \int\frac{1}{u}\, \mathrm{d} u &= \ln|u|+C \\ &= \ln | x^2 + 7x+14 | + C \\ &= \ln ( x^2 + 7x+14 ) + C . \end{align} \]

In this case, we can remove the modulus function from around \( x^2 + 7x+14 \), as this expression is always greater than 0. (This is easily checked by using the discriminant or completing the square).

This is perhaps the simplest substitution to perform. Suppose we have a function that we know how to integrate, but the subject of this function is \(ax + b\) instead of a single variable. This is where the integration by substitution formula \(u = ax + b\) makes things easier. We'll see this in the next few examples.

Find

\[ \int (9x+3)^7 \, \mathrm{d}x.\]

Here, we could either multiply out the binomial, or we can make a substitution. Let's make the substitution of \(u = 9x + 3\), so

\[ \frac{\mathrm{d} u}{\mathrm{d}x} = 9, \]

and so

\[ \mathrm{d}x = \frac{\mathrm{d} u }{9}.\]

This means our integral is now:

\[ \begin{align} \int (9x+3)^7 \, \mathrm{d}x &= \frac{1}{9} \int u^7 \, \mathrm{d}x \\ &= \frac{1}{72}u^8 + C \\ &= \frac{1}{72}(9x+3)^8 + C. \end{align}\]

Find

\[ \int \frac{1}{3x+1} \, \mathrm{d}x.\]

Let's substitute \(u = 3x + 1\), so

\[ \frac{\mathrm{d} u}{\mathrm{d}x} = 3, \]

and so

\[ \mathrm{d}x = \frac{\mathrm{d} u }{3}.\]

Then we can use the substitution.

\[ \begin{align} \int \frac{1}{3x+1} \, \mathrm{d}x &= \frac{1}{3}\int \frac{1}{u} \, \mathrm{d}u \\ &= \frac{1}{3} \ln|u| + C \\ &= \frac{1}{3} \ln|3x+1| + C. \end{align} \]

Find

\[ \int \cos (3x+5) \, \mathrm{d}x.\]

We know how to integrate the cosine function of a single variable, so let's try and convert this expression to that form. Let \(u = 3x + 5\), then

\[ \frac{\mathrm{d} u}{\mathrm{d}x} = 3, \]

and so

\[ \mathrm{d}x = \frac{\mathrm{d} u }{3}.\]

Substituting in, we get

\[ \begin{align} \int \cos (3x+5) \, \mathrm{d}x &= \frac{1}{3} \int \cos u \, \mathrm{d}u \\ &= \frac{1}{3}\sin u + C \\ &= \frac{1}{3}\sin u(3x+5) + C . \end{align}\]

As mentioned earlier, we can get to the idea of integration by substitution by integrating the chain rule. For reference, this is where that leaves us:

\[ \int f'(g(x))g'(x) \, \mathrm{d}x = f(g(x)).\]

The aim is to make the substitution \(u = g (x)\), then

\[ \frac{ \mathrm{d}u}{\mathrm{d}x} = g'(x)\]

and

\[ \mathrm{d}x = \frac{\mathrm{d}u}{g'(x)}.\]

Substituting this in, we get

\[\begin{align} \int f'(g(x))g'(x) \, \mathrm{d}x &= \int f'(u) g'(x)\frac{\mathrm{d}u}{g'(x)} \\ &= \int f'(u)\, \mathrm{d}u \\ &= f(u) \\ &= f(g(x)) \end{align}\]

which is what we wanted.

Note here, that for conciseness, the constant of integration has been excluded.

Integrate

\[ \int 2xe^{x^2}\, \mathrm{d}x.\]

I know that the derivative of \(x^2\) is \(2x\), so that may be a good substitution to make.

If \(u=x^2\), then

\[ \frac{\mathrm{d} u}{\mathrm{d}x} = 2x, \]

and so

\[ \mathrm{d}x = \frac{\mathrm{d} u }{2x}.\]

Substituting this in, we reach

\[ \begin{align} \int 2xe^{x^2}\, \mathrm{d}x &= \int 2xe^u \frac{\mathrm{d} u }{2x} \\ &= \int e^u \, \mathrm{d}u \\ &= e^u + C \\ &= e^{x^2} + C. \end{align} \]

Integrate

\[ \int \frac{\sec^2 x}{1 + \tan x} \, \mathrm{d}x \] by substitution.

Let \(u = 1+ \tan x\), then

\[ \frac{\mathrm{d} u}{\mathrm{d}x} = \sec^2 x, \]

and so

\[ \mathrm{d}x = \frac{\mathrm{d} u}{\sec^2 x},\]

like that

\[\begin{align} \int \frac{\sec^2 x}{1 + \tan x} \, \mathrm{d}x &= \int \frac{\sec^2 x}{u} \frac{\mathrm{d} u }{\sec^2 x} \\ &= \int \frac{1}{u} \, \mathrm{d}u \\ &= \ln|u|+C \\ &= \ln \left| 1+\tan x \right| + C.\end{align} \]

Whilst looking at integration by substitution, there are a couple of key examples to be aware of.

Let us make the substitution \(x = a \tan u\), then

\[ \frac{\mathrm{d} u}{\mathrm{d}x} = a\sec^2 u, \]

and so

\[ \mathrm{d}x = \frac{\mathrm{d} u }{a\sec^2 u}.\]

This then gives

\[ \int \frac{1}{a^2 + x^2} \, \mathrm{d}x = \int \frac{1}{a^2 + a^2\tan^2u} \, \frac{\mathrm{d} u }{a\sec^2 u} .\]

It is worth noting here that

\[ a^2(1+\tan^2 u) = a^2\sec^2 u,\]

as by identity,

\[ 1 + \tan^2 u = \sec^2 u.\]

This means that

\[ \begin{align} \int \frac{1}{a^2 + x^2} \, \mathrm{d}x &= \int \frac{1}{a^2 + a^2\tan^2u} \, \frac{\mathrm{d} u }{a\sec^2 u} \\ & \int \frac{a \sec^2 u}{a^2 \sec^2 u} \, \mathrm{d}u \\ &= \frac{1}{a} \int 1 \, \mathrm{d}u \\ &= \frac{1}{a} u + C \\ &= \frac{1}{a} \arctan \left( \frac{x}{a} \right) + C. \end{align}\]

We can still do integration by substitution when we have a definite integral, however we must remember to change the limits of the integral accordingly. In addition to this, if we change the limits, we no longer need to undo the substitution. This is shown below.

Find

\[ \int_0^5 xe^{x^2} \, \mathrm{d}x .\]

Let \(u = x^2\). Then

\[ \frac{\mathrm{d} u}{\mathrm{d}x} = 2x, \]

and so

\[ \mathrm{d}x = \frac{\mathrm{d} u }{2x}.\]

Then the limits change, to \(0^2 = 0\) and \(5^2 = 25\). This gives

\[ \begin{align} \int_0^5 xe^{x^2} \, \mathrm{d}x &= \frac{1}{2}\int_0^{25} e^u \, \mathrm{d}u \\ &= \frac{1}{2}\left[ e^u\right]_{u=0}^{u=25} \\ &= \frac{e^{25} - 1}{2}. \end{align}\]

- Integration by substitution is the inverse of the chain rule for derivatives.
- When the integral is of the form \[ \int f '(g (x)) g' (x)\, \mathrm{d}x, \] use the substitution \(u = g (x)\).
- When integrating a definite integral, ensure to also use the substitution to shift the limits.
- In an indefinite integral, ensure to undo the substitution, and also include a constant of integration.

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