## Table of Contents

- Change of Variables Integration
- U-Substitution to Find the Definite Integral
- U-Substitution to Find an Indefinite Integral
- U-Substitution Practice
- Lesson Summary

Understand how to use the U substitution method to calculate definite and indefinite integrals. Learn how to incorporate a change of the limits of integration.
Updated: 01/25/2022

- Change of Variables Integration
- U-Substitution to Find the Definite Integral
- U-Substitution to Find an Indefinite Integral
- U-Substitution Practice
- Lesson Summary

Sometimes, in calculus, complex integrands can look impossible to solve, and some integrals are only evaluated with computers. For other integrals, noticing patterns within the integrand can make it easier to solve using the appropriate methods. One such method involves **changing the variables of integration** through a process called **U-substitution**, where **u** is a generic variable that replaces the variable of integration, x. U-substitution skips the derivative chain rule, and if an integrand involves a composite function in the form {eq}f(g(x))g'(x) {/eq}. U-substitution is the best way to make the integral solvable. Figure 1 gives the integrands that have trigonometric substitutions.

A **definite integral**, written as {eq}\int_a^b f(x) dx {/eq}, is an integral with boundaries, where f(x) is some function, a is the lower boundary of the integral, and b is the upper boundary of the integral. When calculating a **U-substitution definite integral**, because the variable of integration changes, it is crucial also to change the boundaries of the integral. The basic method for using U-substitution to perform **definite integral substitution** and appropriately change the bounds of the integral follows these steps:

1) Properly identify that the integral is written as {eq}\int_a^b f(g(x))g'(x) dx {/eq}.

2) Set {eq}u = g(x) {/eq} so that {eq}du = g'(x) dx {/eq}.

3) Put *u* and du in place of g(x) and g'(x)dx so that {eq}\int_a^b f(g(x))g'(x) dx {/eq} becomes {eq}\int_a^b f(u)du {/eq}.

4) The boundaries of integration imply that a and b should be plugged in for x to evaluate the final answer. The variable of integration, however, is not x anymore. Now the variable of integration is *u*, and {eq}u = g(x) {/eq}, so the new boundaries are g(a) and g(b). Note that an alternative method of changing the boundaries of the integral is by keeping the original boundaries and, instead of evaluating the answer in terms of *u*, replace *u* with g(x). Example 3 demonstrates this alternative method.

5) Replace the existing boundaries with the new boundaries, {eq}\int_{g(a)}^{g(b)} f(u)du {/eq}, and evaluate the integral.

1) The integral is in the form {eq}\int_a^b f(g(x))g'(x) dx {/eq} where {eq}g(x) = x^2 {/eq} and g'(x) = 2x.

2) Setting *u* = g(x) gives {eq}u = x^2 {/eq}. Taking the derivative of *u* gives u' = 2xdx.

3) Making the appropriate substitutions into the integral {eq}\int_2^1 2xsin(x^2)dx {/eq} gives {eq}\int_2^1 sin(u)du {/eq}.

4) The boundaries of the integral become {eq}g(2) = 2^2 = 4 {/eq} and {eq}g(1) = 1^2 = 1 {/eq}.

5) Replacing the limits of integration with the new boundaries, {eq}\int_4^1 sin(u)du {/eq}, and evaluating the resulting integral gives:

{eq}- {/eq} the antiderivative of sin(x) is -cos(x) so the integral evaluates to {eq}-cos(u)|_4^1 {/eq}

{eq}- {/eq} evaluating the integral at the boundaries gives -cos(1) + cos(4) = -0.654.

1) The integral is in the form of {eq}\int_a^b f(g(x))g'(x) dx {/eq} where {eq}g(x) = 3x^2 {/eq} and g'(x) = 6x.

2) Setting *u* = g(x) gives {eq}u = 3x^2 {/eq}. Taking the derivative of *u* gives u' = 6xdx.

3) Making the appropriate substitutions into the integral {eq}\int_{9.2}^{12.4} 6xln(3x^2) dx {/eq} gives {eq}\int_{9.2}^{12.4} ln(u) du {/eq}.

4) The boundaries of the integral become {eq}g(9.2) = 3*9.2^2 = 253.92 {/eq} and {eq}g(12.4) = 3*12.4^2 = 461.28 {/eq}.

5) Replacing the limits of integration with the new boundaries, {eq}\int_{253.92}^{461.28} ln(u) du {/eq}, and evaluating the resulting integral gives:

{eq}- {/eq} the antiderivative of ln(u) is {eq}\frac{1}{u} {/eq} so the integral evaluates to {eq}\frac{1}{u}|_{253.92}^{461.28} {/eq}

{eq}- {/eq} evaluating the integral at the boundaries gives {eq}\frac{1}{461.28} - \frac{1}{253.92} = -0.002 {/eq}

1) The integral is not in the form {eq}\int_a^b f(g(x))g'(x) dx {/eq}, and this example will show how to use the integrand to operate U-substitution. Here, {eq}g(x) = 2x^2-1 {/eq} and g'(x) = 4x.

2) Setting *u* = g(x) gives {eq}u = 2x^2-1 {/eq}. Taking the derivative of *u* gives du = 4xdx. This can be solved for dx so that {eq}\frac{du}{4x} = dx {/eq}.

3) Making the appropriate substitutions into the integral {eq}\int_{1}^{\pi} xe^{2x^2-1} dx {/eq} gives {eq}\int_{1}^{\pi} \frac{du}{4x}xe^{u} {/eq}. This is simplified to {eq}\frac{1}{4} \int_{1}^{\pi} e^{u} du {/eq}. Since all of the derivatives of *u* cancel out, there is an extra factor that can be pulled out of the integrand.

4) The boundaries of the integral become {eq}g(1) = 2*1^2-1 = 1 {/eq} and {eq}g(\pi) = 2* \pi^2-1 = 18.74 {/eq}.

5) Replacing the limits of integration with the new boundaries, {eq}\frac{1}{4} \int_{1}^{18.74} e^{u} du {/eq}, and evaluating the resulting integral gives:

{eq}- {/eq} the antiderivative of {eq}e^u {/eq} is {eq}e^u {/eq} so the integral evaluates to {eq}\frac{1}{4} e^u|_1^{18.74} {/eq}

{eq}- {/eq} evaluating the integral at the boundaries gives {eq}\frac{1}{4} (e^{18.74} - e^1) = 3.438*10^7 {/eq}

6) Alternatively:

{eq}- {/eq} {eq}\frac{1}{4} \int_{1}^{\pi} e^{u} du {/eq} evaluates to {eq}\frac{1}{4} e^{u} |_1^{\pi} {/eq}

{eq}- {/eq} replacing *u* with {eq}g(x) = 2x^2-1 {/eq} gives {eq}\frac{1}{4} e^{2x^2-1} |_1^{\pi} {/eq}

{eq}- {/eq} evaluating this gives {eq}\frac{1}{4} (e^{2*{\pi}^2-1} - e^{2*1^2-1}) = \frac{1}{4} (e^{18.74} - e^1) {/eq}

An **indefinite integral** is an integral without boundaries, and indefinite integrals are evaluated with an integration constant, C. During a u-substitution on an indefinite integral, since an indefinite integral has no boundaries, there is no need for **changing limits of integration**. To evaluate an indefinite integral using u-substitution, follow steps 1 - 3 of evaluating a definite integral using u-substitution. After step 3, solve the integral.

1) The integral is in the form {eq}\int f(g(x))g'(x) dx {/eq} where {eq}g(x) = x^2 {/eq} and g'(x) = 2x.

2) Setting *u* = g(x) gives {eq}u = x^2 {/eq}. Taking the derivative of *u* gives u' = 2xdx.

3) Making the appropriate substitutions into the integral {eq}\int 2xsin(x^2)dx {/eq} gives {eq}\int sin(u)du {/eq}.

4) Evaluating the resulting integral and replacing *u* with {eq}g(x) = x^2 {/eq} gives {eq}-cos(x^2) + C {/eq}.

1) The integral is in the form {eq}\int f(g(x))g'(x) dx {/eq} where {eq}g(x) = 3x^2 {/eq} and g'(x) = 6x.

2) Setting *u* = g(x) gives {eq}u = 3x^2 {/eq}. Taking the derivative of *u* gives *u*' = 6xdx.

3) Making the appropriate substitutions into the integral {eq}\int 6xln(3x^2) dx {/eq} gives {eq}\int ln(u) du {/eq}.

4) Evaluating the resulting integral and replacing *u* with {eq}g(x) = 3x^2 {/eq} gives {eq}\frac{1}{3x^2} + C {/eq}.

1) The integral is not in the form {eq}\int f(g(x))g'(x) dx {/eq}, and {eq}g(x) = 2x^2-1 {/eq} and g'(x) = 4x.

2) Setting *u* = g(x) gives {eq}u = 2x^2-1 {/eq}. Taking the derivative of *u* gives du = 4xdx, and {eq}\frac{du}{4x} = dx {/eq}.

3) Substituting the integral and simplifying it gives {eq}\frac{1}{4} \int e^u du {/eq}.

4) Evaluating the resulting integral and replacing *u* with {eq}g(x) = 2x^2-1 {/eq} gives {eq}\frac{1}{4} e^{2x^2-1} + C {/eq}.

Use **change of variables integration** to complete the following practice problems:

** 1) Evaluate {eq}\int x5^{3x^2} dx {/eq}.**

The integrand can be manipulated to use u-substitution. Here, {eq}u = 3x^2 {/eq} and du = 6xdx so {eq}\frac{du}{6x} = dx {/eq}. The resulting integral is {eq}\int \frac{du}{6x} x5^u {/eq}, which simplifies to {eq}\frac{1}{6} \int 5^u du {/eq}. The integral evaluates to {eq}\frac{1}{6} 5^{u-1} {/eq}, which simplifies to {eq}\frac{1}{6} 5^{3x^2-1} + C {/eq}.

** 2) Evaluate {eq}\int_0^{\pi} sin(x) ln(cos(x)) dx {/eq}.**

This integral is in the proper form where *u* = cos(x) and du = -sin(x)dx. Making the appropriate substations into the integral gives {eq}- \int_0^{\pi} ln(u) du {/eq}. The boundary changes, g(0) = 1 and {eq}g(\pi) {/eq} = -1, make the final integral {eq}- \int_1^{-1} ln(u) du {/eq}, which simplifies to {eq}- \frac{1}{u} |_1^{-1} {/eq}. This evaluates to {eq}- \frac{1}{-1} + \frac{1}{1} = 2 {/eq}.

** 3) Evaluate {eq}\int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt(1 + 25x^2) dx {/eq}.**

This integral involves a trigonometric substitution where {eq}a^2 = 1 {/eq} and {eq}b^2 = 25 {/eq} so the substitution {eq}\frac{a}{b} tan( \theta) {/eq} should be used. To find new boundaries, solve +-{eq}\frac{\pi}{2} = \frac{1}{5} tan(\theta) {/eq} for {eq}\theta {/eq}. By using inverse tangent, the new boundaries will be +- 1.444, and the integral becomes {eq}\frac{1}{5} \int_{-1.444}^{1.444} tan(\theta) d \theta {/eq}. The antiderivative of tan(x) is {eq}sec^2(x) {/eq}, so this integral becomes {eq}\frac{1}{5} sec^2(\theta) |_{-1.444}^{1.444} {/eq}. Evaluating this antiderivative gives 0.

In calculus, **changing the variables of integration** is also called **U-substitution**. This method is used for solving integrals whose integrands contain composite functions, and U-substitution is analogous to the derivative chain rule. A **change of variable integral** can greatly help to simplify complex integrands. For example, some radical integrands can be simplified with trigonometric substitutions. The U-substitution method can be used to solve **definite integrals** and **indefinite integrals**.

When solving a **U-substitution definite integral**, it is important to evaluate the integral correctly by either **changing the limits of integration** or by making sure to replace *u* with g(x) before evaluating the final antiderivative. To perform **definite integral substitution** by changing the limits of integration, always identify which function is g(x). Next, evaluate this function at the original boundaries. The results are the new boundaries. When using U-substitution to evaluate an indefinite integral, no change of boundaries is needed, but it is important to include + C, the integration constant.

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- Activities
- FAQs

In this activity, you'll assess your knowledge in using trigonometric substitution to solve definite integrals.

Here are the three cases on how to change variables via trigonometric substitution.

- Case 1: If the integrand has the form sqrt(
*a*^2 - x^2) where*a*is greater than 0, then we introduce a variable theta by letting**x =**. This gives rise to*a*sin(theta)**sqrt(**.*a*^2 - x^2) =*a*cos(theta) - Case 2: If the integrand has the form sqrt(
*a*^2 + x^2) where*a*is greater than 0, then we introduce a variable theta by letting**x =**. This gives rise to*a*tan(theta)**sqrt(**.*a*^2 + x^2) =*a*sec(theta) - Case 3: If the integrand has the form sqrt(x^2 -
*a*^2) where*a*is greater than 0, then we introduce a variable theta by letting**x =**. This gives rise to*a*sec(theta)**sqrt(x^2 -**.*a*^2) =*a*tan(theta)

- Blank paper
- Pencil

Before you answer the practice problems, let us first look at the steps in evaluating the expression below:

**Step 1:** Determine the case where the integrand sqrt(*sqrt(5)*^2 - x^2) belongs. By inspection, one can see that the integrand follows case 1, where a = sqrt(5).

**Step 2:** Now, let x = sqrt(5)sin(theta) and solve for *dx *in terms of *d(theta)*.

**Step 3:** It follows from case 1 that the integrand will be equal to

**Step 4:** Rewrite and evaluate the integral using the equations obtained in Step 2 and 3.

**Step 5:** Change the limits by using the equation in Step 2,

**Step 6:** Evaluate Step 4 using the new limits.

Therefore, evaluating the integral via trigonometric substitution yields **0.1**.

Now let's see if you understand the steps outlined above. With your pen and paper, provide clear solutions and answers to the given problems.

To find the upper and lower boundaries of the new integral in the U-substitution, identify which function is g(x). Evaluate g(x) at each boundary, and the results are the new lower and upper boundaries.

To change integration variables, first identify one function, g(x), in the integrand to be 'u'.Take the derivative of this function, du = g'(x)dx. Replace g(x) with u and g'(x)dx with du. Finally, integrate over u.

In calculus, U-substitution is a method to calculate integrals whose integrands contain composite functions. U-substitution is the integral version of the derivative chain rule.

Yes. U-substitution can solve definite integrals as long as an appropriate change of boundaries is made. U-substitution can also solve indefinite integrals with no change of limits.

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