In this explainer, we will learn how to use integration by substitution for definite integrals.

Let us first recall how to integrate by substitution by going through a quick example. Let

We compute this integral by making the substitution . We have that and so

In this form, the integral is straightforward to compute:

Finally, we substitute back in:

In general, if we have a function of a function and an integral of the form , then the substitution yields , and so

This technique works for definite integrals just as it does for indefinite ones; however, when we make a substitution, the limits of integration need to change.

### Formula: Integration by Substitution for Definite Integrals

If is a continuous function and is differentiable with continuous derivative, then where .

Let us look at an example of using substitution to evaluate a definite integral.

### Example 1: Finding the Value of a Definite Integral Using Substitution

Use the substitution to evaluate the definite integral .

### Answer

Recall that if we are integrating a function of the form , then we can make the substitution :

Observe that after substituting the limits of integration change from and to and .

We want to evaluate . Taking , we have the following:

- the limits of integration and

Thus,

If we are not told which substitution to use, we will have to choose a suitable substitution ourselves.

### Example 2: Finding the Value of a Definite Integral Using Substitution

Find to the nearest thousandth.

### Answer

Recall that if we have a function of the form , then we can make the substitution :

Observe that after substituting the limits of integration change from and to and .

In making a substitution, we are looking for two things. The first thing is an expression, call it , that appears in the integrand and whose derivative, , also appears in the integrand. It is worth remembering that if is linear in (i.e., is of the form ), then its derivative is a constant; . This means that a linear substitution can always be made.

The second thing we are looking for is a substitution that will make the integral easier to evaluate. When the integrand contains a function of a function, the inner function is often a good candidate for substitution.

These two considerations suggest that would be a good substitution to try. Indeed, taking , we have the following:

- the limits of integration and

Thus,

Therefore, the value of our definite integral to the nearest thousandth is 21.067.

The method of substitution is often an effective way of tackling trigonometric integrals, as we will see in the next example.

### Example 3: Finding the Value of a Definite Integral Using Substitution

Find .

### Answer

Recall that if we have a function of the form , then we can make the substitution :

Observe that after substituting the limits of integration change from and to and .

In making a substitution, we are looking for an expression, call it , that appears in the integrand and whose derivative, , also appears in the integrand. Now, trigonometric functions differentiate to other trigonometric functions. For this reason, when presented with an integral involving multiple trigonometric functions, it is worth checking to see if any of them are the derivatives of other ones.

We want to evaluate . Taking , we have the following:

- the limits of integration and

Thus,

Recall that the derivative of is . This means that quotients with in the numerator are often good candidates for integration by substitution, as in the next example.

### Example 4: Finding the Value of a Definite Integral Using Substitution

Determine .

### Answer

Recall that if we have a function of the form , then we can make the substitution :

Observe that after substituting the limits of integration change from and to and .

Since our integrand in this case consists of a fraction with in the numerator, is an obvious candidate for substitution. Indeed, taking , we have the following:

- the limits of integration and

Thus,

Let us look at a more complicated example of using substitution to evaluate a definite integral involving the natural logarithm.

### Example 5: Finding the Value of a Definite Integral Using Substitution

Evaluate .

### Answer

Recall that if we have a function of the form , then we can make the substitution :

Observe that after substituting the limits of integration change from and to and .

Although the integrand in this example looks rather complicated, we should immediately notice that it contains a natural logarithm and a fraction whose denominator is closely related to the argument of that logarithm. By using the chain rule, we get

This suggests that might be an extremely good candidate for substitution. However, observe that the constant term in differentiates to 0, so we may as well take . With , we have the following:

- the limits of integration and

Thus,

Finally, let us look at an example combining an exponential function and trigonometric functions.

### Example 6: Finding the Value of a Definite Integral Using Substitution

Evaluate .

### Answer

Recall that if we have a function of the form , then we can make the substitution :

Observe that after substituting the limits of integration change from and to and . Consider the expression

Notice the presence of a sine function in the exponent and right next to it its derivative, a cosine. Indeed, with , we have , and so

We have and . Therefore,

Let us finish by recapping a few important concepts from this explainer.

### Key Points

- We can identify situations where a substitution can be used to simplify a definite integral.
- In particular, we can recognize expressions of the form which are the product of a function of a function and the derivative of the inner function.
- In this situation, we can make the substitution ; that is,
- We can evaluate such definite integrals, taking care that when we make a substitution , the limits of integration change: