# Introduction to Integration – Integration by U-Substitution – Change of Variable Method – the Reverse Chain Rule

In a previous section, we learned about the concept of the indefinite integral and how to find the antiderivative of simple, common functions using the table of integrals. For more complicated functions though, using basic antidifferentiation techniques or a table of integrals will not work. This section will cover one of the more advanced methods for finding the indefinite integral for more complex functions: U- substitution.

U-substitution helps us to find the integral of functions y = f(g(x))g’(x) in which outside function u’ = g’(x) is the derivative of the inside function u = g(x).

Before we consider integration by U-substitution, we need to determine if we can perform U-substitution on a function:

Example 1: Can we perform U-substitution on f(x) = 2xcos(x2)?

Solution: First, we need to determine the inside function.

The inside function is u = x2.

Next, we need to find du and see if it matches the outside function:

du =(x2)’ = 2x

du= 2x is located in the function. Therefore, U-substitution can be performed. u and du can be substituted into the original function to obtain:

f(x) = du cos(u) or cos(u)du

Example 2: Can we perform U-substitution on

Solution: First, we need to determine the inside function.

The inside function is u = x2+1

Next, we need to find du and see if it matches the outside function.

du = (x2+1)’ = 2x

We can divide each side by 2 to make du match the term in the numerator:

½ du = x

Therefore, U-substitution can be performed. u and ½du can be substituted into the original function to obtain:

Example 3: Can we perform U-substitution on f(x) = 6x2(2x4+5)6?

Solution: First, we need to determine the inside function.

The inside function is u = 2x4+5

Next, we need to find du and see if it matches the outside function.

du = (2x4+5)’ = 8x3

We see that du = 8x3is not equal to 6x2as seen in the function. Therefore, U-substitution cannot be performed.

Example 4: Can we perform U-substitution on f(x) = 3x(2x+7)6?

Solution: First, we need to determine the inside function.

The inside function is u = 2x+7

Next, we need to find du and see if it matches the outside function.

du = (2x+7)’ = 2

We see that du =2is not equal to 3x as seen in the function. Therefore, U-substitution cannot be performed.

Example 5: Can we perform U-substitution on f(x) = cos(3x+2)?

Solution: First, we need to determine the inside function.

The inside function is u = 3x+2

Next, we need to find du and see if it matches the outside function.

du = u ‘= 3

We see that du = 3, which does not match anything in the original function. However, we can divide both sides by 3 to obtain:

⅓ du = 1

However, we do not see a 1 in the function. Although this is true, the function f(x) cos(3x+2) can be written as cos(3x+2) * 1 (they both mean the same thing). Therefore, 1/3 du is indeed in the function. Therefore, U-substitution can be performed.

u and du can be substituted into the original function to obtain:

f(x) = cos(3x+2) * 1

f(x) = cos(u) * ⅓ du

In each composite function, we determined the inside function, u, and took the derivative of it, du, which is the outside function. If u = g(x), and du = g’(x), then our composite function is

y = f(g(x))*g‘(x) = f(u) * u’

This looks like the chain rule of differentiation.

Therefore, if we are integrating, then we are essentially reversing the chain rule. Hence, U-substitution is also called the ‘reverse chain rule’.

Although the notation is not exactly the same, the relationship is consistent.

Therefore, integration by U-substitution can be written as the following:

 Formula: U – Substitution If u = g(x), and du = g’(x), then the integral of the composite function f(g(x))*g‘(x) is Guidelines for applying U-substitution: Determine your inside function u = g(x). Note: The inside function will typically be in parenthesis or under a square root (not always). u is typically the more ‘complicated’ function in the integrand. Select your u in which that, if you take the derivative of u, du will match the outside function. Take the derivative of u to obtain du = g’(x)dx. Note: Make sure the du term matches the outside function. If it doesn’t match the outside function right away, then you need to perform some arithmetic to the du to obtain the outside function. Rewrite the integral in terms of u and du. Find the resulting integral in terms of u. After integrating, replace u = g(x) inside the antiderivative. Bonus: Check your work by differentiating the anti-derivative using the chain rule. Note: This makes sense, because U-substitution is the reverse chain rule, so performing the chain rule for differentiation on the antiderivative should give us the original function f(x). (Remember the Fundamental Theorem of Calculus).

Example 1: Suppose you have an indefinite integral

Find the antiderivative by using the U-substitution method.

Solution:

Determine your inside function u = g(x).

We see that x2 is inside the sine function. Therefore, let’s select u = x2

Take the derivative of u to obtain du = g’(x)dx.

u = x2

du = 2xdx

Rewrite the integral in terms of u and du.

Replace x2with u, and 2xdx with du.

Find the resulting integral in terms of u.

Let’s integrate our function, treating u as a variable:

F(x) = ∫ Sin (u) d (u)

F(x) = – Cos (u) + C, as the integral of sine is –cosine,

After integrating, replace u = g(x) inside the antiderivative.

Last, put the u = x2 back to ‘u’,

F(x) = – Cos (x2) + C

Bonus: Check your work by differentiating the anti-derivative using the chain rule.

The outside function is f(x) = -cos(x), so the derivative is f’(x) = sin(x). The inside function is u = g(x) = x2, so u’ = g’(x) = 2x. Performing the chain rule:

F’ (x) = f’(g(x) )* g’(x) = sin (x2)*2x + 0 = sin (x2)*2x

Since the result of F’(x) matches the integrand, the U-substitution was performed correctly.

Example 2: Find the result of F(x) = ∫ (2x + 5)dx.

Solution:

Determine your inside function u = g(x).

We see that 2x + 5 is inside parenthesis. Therefore, let’s select u = 2x + 5

Take the derivative of u to obtain du = g’(x)dx.

u = 2x + 5

du = 2dx

du doesn’t exactly match any function in the integrand.

Therefore, with du = 2xdx, let’s divide both sides by 2:

½du = dx

Therefore, ½du = dx and u =2x+5.

Rewrite the integral in terms of u and du.

Replace ½du = dx and u =2x+5 inside the integrand:

Factor the ½ from the integrand:

∫ (u)½du = ½ ∫ (u)du

Find the resulting integral in terms of u.

Let’s integrate our function, treating u as a variable. Applying the power rule of integration,

In this example, we will not check our work by differentiating, but if this were to be done, F’(x) will be equal to the original expression in the integrand.

Let’s try an example for the definite integrals:

Example 3: Find the area under the curve of f(x) = cos(x)esin(x) from x = 0 to x = 1.

Solution:

We are essentially finding the definite integral of f(x) on the interval [0,1]:

Determine your inside function u = g(x).

Between sin(x), cos(x), and ex, we need to select a function which, if the derivative is taken, will equal another function in the integrand.

Let’s try u = ex:

u = ex

du = exdx

Since du = exdx does not equal any function in the integrand, a new u needs to be selected.

Let’s try u = cos(x)

du = -sin(x)dx

Even though we have sin(x) in the exponent, du will not be the exponent because the exponent does not have the differential dx.

Let’s try u = sin(x)

du = cos(x) dx

In this case, cos(x)dx exists in the integrand. Therefore, let’s use u = sin(x) and du = cos(x)dx

Take the derivative of u to obtain du = g’(x)dx.

As already determined in the previous step,

u = x2

du = 2xdx

Rewrite the integral in terms of u and du.

Replace sin(x) with u, and cos(x)dx with du.

Find the definite integral:

F(b) – F(a) = e1-e0= e-1

Without considering the limits of integration, the indefinite integral would be:

F(x) = eudu = eu+ C

Substituting u = sin(x) back into the antiderivative,

F(x) = esin(x)+ C

Bonus: Check your work by differentiating the anti-derivative using the chain rule.

The outside function is f(x) = e(x) , so the derivative is f’(x) = 1*e(x)= ex. The inside function is u = g(x) = sin(x) , so u’ = g’(x) = cos(x). Perform the chain rule,

F’ (x) = f’(g(x) )* g’(x) =esin(x)*cos(x)= esin(x)cos(x)

Since the result of F’(x) matches the integrand, the U-substitution was performed correctly.

Example 4: Find the antiderivative of

Solution: Determine your inside function u = g(x).

Between 3x and 6x2+5, we need to select a function which, if the derivative is taken, will equal another function in the integrand.

Let’s try u = 3x:

du = 3dx

Since du = 3dx does not equal any function in the integrand, a new u needs to be selected.

Let’s try u = 6x2+5

du = 12xdx

Our du equation doesn’t match the 3xdx in the integrand. However, we can fix this in one of two ways:

1. Factor the 3 from the integrand to obtain:

Then divide the equation of du = 12xdx by 12 on both sides to obtain:

1/12 du = x dx

Now the du equation matches the outside function in the integrand.

1. Divide each side of the du equation by 4 to obtain:

¼du = 3xdx

Now the du equation matches the outside function in the integrand.

Let’s use the first option.

Take the derivative of u to obtain du = g’(x)dx.

As already determined in the previous step,

u = 6x2+5

1/12 du = xdx

Rewrite the integral in terms of u and du.

Replace 6x2+5 with u and xdx with du.

Example 5: Find the antiderivative of F(x) =∫ x(4x-1)4dx

Solution: Determine your inside function u = g(x).

We see that 4x – 1 is the inside function. Let’s use u = 4x – 1

Take the derivative of u to obtain du = g’(x)dx.

u = 4x – 1

du = 4 dx

However, we do not see 4 dx in the integrand. However, we can change this by either dividing by 4 on both sides or multiplying by ¼ on both sides:

du = 4 dx

¼* du = ¼ * 4 dx

¼du = dx

Now the du equation matches the differential in the integrand.

However, we still need to replace the x in the front of the integrand. We need to go back to our u equation and solve for x:

u = 4x -1

u + 1 = 4x

u/4 + 1/4 = x

Therefore, the integrand can be replaced with the following:

u = 4x – 1

¼ du = dx

u/4 + 1/4 = x

Rewrite the integral in terms of u and du.

Replace 4x – 1 = u, dx = ¼ du, and x = u/4 + ¼ inside the integrand:

 Summary of Section U-substitution helps us to find the integral of composite functions in which the function on the outside ‘du’ is the derivative of the inside function ‘u’. U-substitution (the reverse chain rule) and the chain rule of integration are the inverse operations of each other. If part of the integrand is a composition of functions, f(g(x)), then try setting u = g(x ), the ‘inner’ function. If we have a function in the form y = f(g(x))*g’(x), we can substitute u = g(x) and du = g’(x) into the formula. Then the integral of a composite function: