Introduction

This topic introduces u-substitution, the most important integration technique in calculus. The essential knowledge tells us that substitution of variables is a technique for finding antiderivatives and for a definite integral, substitution of variables requires corresponding changes to the limits of integration. Mastering this technique is crucial for evaluating integrals that cannot be solved by basic formulas alone.

Understanding U-Substitution

The Fundamental Idea

U-substitution reverses the chain rule. When we have an integral containing a function and its derivative, we can simplify by substituting.

Chain Rule: $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$

Reversing it: $\int f^{\prime }(g(x))\cdot g^{\prime }(x),dx=f(g(x))+C$

U-Substitution format: If $u=g(x)$, then $du=g^{\prime }(x),dx$

$\int f^{\prime }(u),du=f(u)+C$

When to Use Substitution

Look for integrals where:

Example 1: Evaluate $\int 2x(x^2+1{)}^5,dx$.

Notice: $(x^2 + 1)$ is inside the power, and its derivative $2x$ appears!

Let $u=x^2+1$ Then $du=2x,dx$

Substitute: $\int (x^2+1{)}^5\cdot 2x,dx=\int u^5,du$

Integrate: $=\frac{u^6}{6}+C$

Substitute back: $=\frac{(x^2+1{)}^6}{6}+C$

Basic Substitution Patterns

Linear Inside Functions

Example 2: Evaluate $\int (3x+2{)}^7,dx$.

Let $u=3x+2$ Then $du=3,dx$, so $dx=\frac{1}{3}du$

Substitute: $\int (3x+2{)}^7,dx=\int u^7\cdot \frac{1}{3}du$

$=\frac{1}{3}\int u^7,du$

$=\frac{1}{3}\cdot \frac{u^8}{8}+C$

$=\frac{(3x+2{)}^8}{24}+C$

Adjusting Constants

Sometimes the derivative doesn't appear exactly—we need to adjust by a constant.

Example 3: Evaluate $\int x(x^2+5{)}^4,dx$.

Let $u=x^2+5$ Then $du=2x,dx$, so $x,dx=\frac{1}{2}du$

Substitute: $\int x(x^2+5{)}^4,dx=\int (x^2+5{)}^4\cdot x,dx$

$=\int u^4\cdot \frac{1}{2},du$

$=\frac{1}{2}\int u^4,du$

$=\frac{1}{2}\cdot \frac{u^5}{5}+C$

$=\frac{(x^2+5{)}^5}{10}+C$

Trigonometric Substitutions

Sine and Cosine Patterns

Example 4: Evaluate $\int {\sin }^3(x)\cos (x),dx$.

Let $u=\sin (x)$, then $du=\cos (x),dx$

Substitute: $‌\displaystyle\int \sin^3(x) \cos(x) , dx = \displaystyle\int u^3 , du$

$=\frac{u^4}{4}+C$

$=\frac{{\sin }^4(x)}{4}+C$

Example 5: Evaluate $\int \tan (x),dx$.

Rewrite: $\int \tan (x),dx=\int \frac{\sin (x)}{\cos (x)},dx$

Let $u=\cos (x)$ Then $du=-\sin (x),dx$, so $\sin (x)dx=-du$

Substitute: $\int \frac{\sin (x)}{\cos (x)}dx=\int \frac{-du}{u}$

$=-\ln |u|+C$

$=-\ln |\cos (x)|+C$

Or equivalently: $=\ln |\sec (x)|+C$

Tangent and Secant Patterns

Example 6: Evaluate $\int {\sec }^2(3x),dx$.

Let $u=3x$, then $du=3,dx$, so $dx=\frac{1}{3}du$

Substitute: $\int {\sec }^2(3x),dx=\int {\sec }^2(u)\cdot \frac{1}{3},du$

$=\frac{1}{3}\tan (u)+C$

$=\frac{1}{3}\tan (3x)+C$

Exponential and Logarithmic Substitutions

Exponential Functions

Example 7: Evaluate $\int x{e}^{x^2},dx$.

Let $u=x^2$ Then $du=2x,dx$, so $x,dx=\frac{1}{2}du$

Substitute: $\int x{e}^{x^2},dx=\int e^{x^2}\cdot x,dx$

$=\int e^u\cdot \frac{1}{2},du$

$=\frac{1}{2}{e}^u+C$

$=\frac{1}{2}{e}^{x^2}+C$

Example 8: Evaluate $\int \frac{e^{2x}}{1+e^{2x}},dx$.

Let $u=1+e^{2x}$ Then $du=2e^{2x},dx$, so $e^{2x},dx=\frac{1}{2}du$

Substitute: $\int \frac{e^{2x}}{1+e^{2x}},dx=\int \frac{1}{u}\cdot \frac{1}{2},du$

$=\frac{1}{2}\ln |u|+C$

$=\frac{1}{2}\ln |1+e^{2x}|+C$

Since $1+e^{2x}>0$: $=\frac{1}{2}\ln (1+e^{2x})+C$

Logarithmic Integrands

Example 9: Evaluate $\int \frac{2x}{x^2+1},dx$.

Let $u=x^2+1$ Then $du=2x,dx$

Substitute: $\int \frac{2x}{x^2+1},dx=\int \frac{du}{u}$

$=\ln |u|+C$

$=\ln |x^2+1|+C$

Since $x^2+1>0$: $=\ln (x^2+1)+C$

Definite Integrals with Substitution

Method 1: Substitute Back to x

Example 10: Evaluate ${\int }_0^{2}x\sqrt{x^2+1},dx$.

Let $u=x^2+1$, then $du=2x,dx$, so $x,dx=\frac{1}{2}du$

Substitute: $\int x\sqrt{x^2+1},dx=\int \sqrt{u}\cdot \frac{1}{2},du$

$=\frac{1}{2}\int u^{1/2},du$

$=\frac{1}{2}\cdot \frac{2u^{3/2}}{3}+C$

$=\frac{u^{3/2}}{3}+C$

$=\frac{(x^2+1{)}^{3/2}}{3}+C$

Now evaluate: ${\int }_0^{2}x\sqrt{x^2+1},dx={\left[\frac{(x^2+1{)}^{3/2}}{3}\right]}_0^2$

$=\frac{(4+1{)}^{3/2}}{3}-\frac{(0+1{)}^{3/2}}{3}$

$=\frac{5^{3/2}}{3}-\frac{1}{3}$

$=\frac{5\sqrt{5}-1}{3}$

Method 2: Change the Limits

Example 11: Evaluate ${\int }_0^{2}x\sqrt{x^2+1},dx$ by changing limits.

Let $u=x^2+1$ Then $du=2x,dx$, so $x,dx=\frac{1}{2}du$

Change limits: When $x=0$: $u=0^2+1=1$ When $x=2$: $u=2^2+1=5$

Substitute with new limits: ${\int }_0^{2}x\sqrt{x^2+1},dx={\int }_1^5\sqrt{u}\cdot \frac{1}{2},du$

$=\frac{1}{2}{\int }_1^5{u}^{1/2},du$

$=\frac{1}{2}{\left[\frac{2u^{3/2}}{3}\right]}_1^5$

$=\frac{1}{3}{\left[u^{3/2}\right]}_1^5$

$=\frac{1}{3}(5^{3/2}-1^{3/2})$

$=\frac{5\sqrt{5}-1}{3}$

Multiple Substitution Attempts

Example 12: Evaluate $\int \frac{x}{(x+1{)}^2},dx$.

Method 1: Let $u=x+1$, so $x=u-1$ and $dx=du$

$\int \frac{x}{(x+1{)}^2},dx=\int \frac{u-1}{u^2},du$

$=\int \left(\frac{1}{u}-\frac{1}{u^2}\right),du$

$=\ln |u|+\frac{1}{u}+C$

$=\ln |x+1|+\frac{1}{x+1}+C$

Recognizing When Substitution Won't Work

When the Derivative Doesn't Appear

Example 13: Can we use substitution for $\int x^2\sin (x),dx$?

If we try $u=\sin (x)$, then $du=\cos (x),dx$, but $\cos (x)$ doesn't appear.

If we try $u=x^2$, then $du=2x,dx$, but we still have $\sin (x)$ remaining.

This integral requires integration by parts (not covered in AB).

When to Use Other Techniques

Some integrals require:

Example 14: For $\int {\sin }^2(x),dx$, use the identity ${\sin }^2(x)=\frac{1-\cos (2x)}{2}$

$\int {\sin }^2(x),dx=\int \frac{1-\cos (2x)}{2},dx$

$=\frac{1}{2}\int (1-\cos (2x)),dx$

$=\frac{1}{2}\left(x-\frac{\sin (2x)}{2}\right)+C$

$=\frac{x}{2}-\frac{\sin (2x)}{4}+C$

Common Substitution Patterns Summary

Quick Reference Guide

$\begin{array}{ccc}\text{Integrand Pattern}& \text{Substitution}& \text{Note}\\ f(g(x))\cdot g^{\prime }(x)& u=g(x)& \text{Exact chain rule reversal}\\ (ax+b{)}^n& u=ax+b& \text{Linear inside function}\\ x(x^2+c{)}^n& u=x^2+c& \text{Adjust for constant}\\ {\sin }^n(x)\cos (x)& u=\sin (x)& \text{Or vice versa (for }{\cos }^n(x)\sin (x)\text{, let }u=\cos (x)\text{)}\\ \tan (x)& u=\cos (x)& \text{Results in }-\ln \\ x{e}^{x^2}& u=x^2& \text{Exponential composition}\\ \frac{f^{\prime }(x)}{f(x)}& u=f(x)& \text{Results in }\ln \end{array}$

Problem-Solving Strategy

Step-by-Step Approach

Practice Section