Integration by substitution

Substitution is a very powerful method of integration. While integrating, we replace part of the function with another expression, integrate, and then substitute back.

The basic formula

Let f be a function and F its antiderivative (a function whose derivative is the original function; at A level simply “an integral” of it) on the interval J. Let ϕ be a real function that is differentiable on the interval I and satisfies ϕ(I) ⊆ J. Then

$$\int f(\varphi (t))\cdot {\varphi }^{\prime }(t)\text{ d}t=F(\varphi (t)).$$

How did we get from the left-hand side to the right-hand side?

We will use this theorem by choosing the substitution x = ϕ(t) and changing the variable of integration – from dt we get dx. The derivative of x with respect to t is $\frac{dx}{dt}={\varphi }^{\prime }(t)$, so dx = ϕ′(t) dt. We rewrite the integral on the left by replacing ϕ(t) with x and the whole product ϕ′(t) dt with dx:

$$\int f(\varphi (t))\cdot {\varphi }^{\prime }(t)\text{ d}t=\int f(x)\text{ d}x$$

Since F is an antiderivative of f by assumption, we can replace the whole integral with F:

$$\int f(x)\text{ d}x=F(x)$$

Finally, we reverse the substitution: instead of x we write the original ϕ(t):

$$F(x)=F(\varphi (t))$$

That gives us the equality from the start:

$$\int f(\varphi (t))\cdot {\varphi }^{\prime }(t)\text{ d}t=\dots =F(\varphi (t)).$$

This is roughly how we will use the theorem.

First example

Work out this integral:

$$\int \sin 6t\text{ d}t$$

We know the formula for the integral of sin x (or here sin t), but we don’t know a formula for sin 6t. So we use a substitution. We will substitute x = 6t. By the formula above,

$$\int f(\varphi (t))\cdot {\varphi }^{\prime }(t)\text{ d}t=\int f(x)\text{ d}x,$$

where x = ϕ(t). In our case f = sin and ϕ(t) = 6t. So by the formula we have

$$\int (\sin 6t)\cdot (6t{)}^{\prime }\text{ d}t=\int \sin x\text{ d}x,$$

where x = 6t. The problem is that we want the integral of sin 6t, not of (sin 6t) · (6t)′. What can we do about this? Let’s try differentiating the function 6t. We get:

$$\int (\sin 6t)\cdot 6\text{ d}t=\int \sin x\text{ d}x,$$

The number six is a constant, so we can take it out in front of the whole integral:

$$6\cdot \int (\sin 6t)\text{ d}t=\int \sin x\text{ d}x,$$

Now we divide the whole equation by six:

$$\int (\sin 6t)\text{ d}t=\frac{1}{6}\cdot \int \sin x\text{ d}x,$$

On the left-hand side we now have the original function that we wanted to integrate at the start. All that remains is to work out the integral on the right-hand side. The integral of sin x is −cos x:

$$\int (\sin 6t)\text{ d}t=\frac{1}{6}\cdot (-\cos x),$$

And we substitute 6t back in for x:

$$\int (\sin 6t)\text{ d}t=-\frac{1}{6}\cdot \cos 6t+c,$$

And that is the final result.

First example, made simpler

The previous method was correct, but substitutions are usually written a little differently. Here is the question again:

$$\int \sin 6t\text{ d}t$$

We write the substitution in brackets after the integral like this:

$$\int \sin 6t\text{ d}t=\left[\begin{array}{c}x=6t\\ dx=6dt\end{array}\right]$$

On the first line we write our substitution, that is x = 6t. Now we differentiate both sides: the left side with respect to x and the right side with respect to t. On the second line of the first column we will always have dx, because the derivative of x is always 1. So dx just shows that we differentiated with respect to x.

We differentiate the right-hand side in the usual way with respect to t. We get (6t)′ = 6 and add dt, because we differentiated with respect to t. To make our work easier, we can also write dt on its own on the next line:

$$\int \sin 6t\text{ d}t=\left[\begin{array}{c}x=6t\\ dx=6dt\\ \frac{dx}{6}=dt\end{array}\right]$$

In the next step we carry out the substitution itself. What do we replace with what? We want x in place of 6t, so that is the first substitution. Because we are putting the new variable x into the function, we must change dt as well. In place of dt we therefore substitute the value we got on the third line, that is dx/6.

To recap: in place of 6t we write x (first line) and in place of dt we write dx/6 (third line). We get:

$$\int \sin 6t\text{ d}t=\left[\begin{array}{c}x=6t\\ dx=6dt\end{array}\right]=\int \sin x\frac{dx}{6}$$

From here the method is completely routine. We take the six out in front of the integral:

$$\int \sin x\frac{dx}{6}=\frac{1}{6}\cdot \int \sin xdx$$

We integrate the sine:

$$\frac{1}{6}\cdot \int \sin xdx=\frac{1}{6}\cdot (-\cos x)$$

And we substitute 6t back in for x:

$$\frac{1}{6}\cdot (-\cos x)=-\frac{1}{6}\cdot \cos 6t+c$$

Second example

Work out this integral using substitution:

$$\int \tan t\text{ d}t$$

You could probably find the integral of tan in a table of standard integrals, but we can just as well work it out ourselves. First we write tan as the quotient sin/cos:

$$\int \tan t\text{ d}t=\int \frac{\sin t}{\cos t}\text{ d}t$$

We choose the substitution x = cos t. We get:

$$\int \frac{\sin t}{\cos t}\text{ d}t=\left[\begin{array}{c}x=\cos t\\ dx=-\sin t\text{ d}t\end{array}\right]$$

Now we make dt the subject:

$$dt=-\frac{dx}{\sin t}$$

Now we can substitute into the integral. In place of cos t we write x and in place of dt we write −dx/sin t.

$$\int \frac{\sin t}{\cos t}\text{ d}t=\left[\begin{array}{c}x=\cos t\\ dx=-\sin t\text{ d}t\end{array}\right]=\int -\frac{\sin t}{x}\cdot \frac{dx}{\sin t}$$

A small problem is that the old variable of integration t is still in the expression, inside the sine. Fortunately, the two sines cancel and we are left with:

$$\int -\frac{\sin t}{x}\cdot \frac{dx}{\sin t}=\int -\frac{dx}{x}$$

Let’s rewrite this a little to get a nicer expression. We move dx to the right of the fraction and write a one in the numerator. We move the minus sign in front of the integral:

$$\int -\frac{dx}{x}=-\int \frac{1}{x}\text{ d}x$$

We already know the integral of 1/x: it is a logarithm. So we get:

$$-\int \frac{1}{x}\text{ d}x=-\ln |x|$$

Finally, we substitute the original cos t back in for x.

$$-\int \frac{1}{x}\text{ d}x=-\ln |x|=-\ln |\cos t|+c$$

Third example

Work out this integral using substitution:

$$\int x\cdot {\sin }^{3}t\text{ d}t$$

It may confuse you that the integrand contains two variables: x and t. That doesn’t matter at all, because dt tells us that we are integrating with respect to t, so x behaves like a constant. So let’s not panic and simply move x in front of the integral:

$$\int x\cdot {\sin }^{3}t\text{ d}t=x\cdot \int {\sin }^{3}t\text{ d}t$$

We don’t have many formulae for sin³, but there is something for sin², so we rewrite the function like this:

$$x\cdot \int {\sin }^{3}t\text{ d}t=x\cdot \int {\sin }^{2}t\cdot \sin t\text{ d}t$$

Now we rewrite sin² using the identity sin²t = (1 − cos²t). We get the integral:

$$x\cdot \int {\sin }^{2}t\cdot \sin t\text{ d}t=x\cdot \int (1-{\cos }^{2}t)\cdot \sin t\text{ d}t$$

Now we substitute. The letter x is already taken, so we use y, for example. We choose y = cos t

$$x\cdot \int (1-{\cos }^{2}t)\cdot \sin t\text{ d}t=\left[\begin{array}{c}y=\cos t\\ dy=-\sin tdt\end{array}\right]$$

We make dt the subject:

$$dt=-\frac{dy}{\sin t}$$

And we substitute into the integral:

$$x\cdot \int (1-{\cos }^{2}t)\cdot \sin t\text{ d}t=\left[\begin{array}{c}y=\cos t\\ dy=-\sin tdt\end{array}\right]=-x\cdot \int (1-y^2)\cdot \sin t\cdot \frac{dy}{\sin t}$$

Again the sines cancel nicely:

$$-x\cdot \int (1-y^2)\cdot \sin t\cdot \frac{dy}{\sin t}=-x\cdot \int (1-y^2)\cdot dy$$

We can split this integral into two integrals:

$$-x\cdot \int (1-y^2)\cdot dy=-x\cdot (\int 1\text{ d}y-\int y^2\text{ d}y)$$

These are both standard integrals:

$$-x\cdot (\int 1\text{ d}y-\int y^2\text{ d}y)=-x\cdot (y-\frac{y^3}{3})$$

Now we substitute the original cos t back in for y:

$$-x\cdot (y-\frac{y^3}{3})=-x\cdot (\cos t-\frac{{\cos }^{3}t}{3})+c$$

That is the final result.