The Substitution Rule for MATH 122
Exam Relevance for MATH 122
Primary integration technique tested. Expect standalone u-substitution problems on every exam.
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Understanding U-Substitution
Not every integral can be solved by just looking at a table of formulas. Sometimes you'll see something like $\int 2x(x^2 + 1)^5 \, dx$ and think "I have no idea what function has this as its derivative." The basic rules (power rule, trig integrals, etc.) don't directly apply because the expression is too complicated.
That's where u-substitution comes in. It's a technique that transforms a messy integral into a simpler one that you can recognize.
The key insight: u-substitution is the reverse of the chain rule. When you see a composite function inside an integral, u-substitution helps you simplify it by temporarily replacing part of the expression with a single variable $u$.
The Core Idea
Remember the chain rule for derivatives: $$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$
U-substitution reverses this. If you spot something that looks like $f'(g(x)) \cdot g'(x)$, you can integrate it as $f(g(x)) + C$.
The Method
- Choose $u$: Pick an "inside" function (usually something inside parentheses, under a root, or in an exponent)
- Find $du$: Differentiate $u$ to get $du = u'(x) \, dx$
- Substitute: Replace all $x$'s with $u$'s (the original variable should disappear completely)
- Integrate: Solve the simpler integral in terms of $u$
- Back-substitute: Replace $u$ with the original expression
How to Choose $u$
Look for these patterns:
| What you see | Choose $u$ as |
|---|---|
| $(...)^n$ where the derivative of inside is present | The inside part |
| $e^{(...)}$ | The exponent |
| $\ln(...)$ | The argument of ln |
| $\sin(...)$ or $\cos(...)$ | The argument |
| $\sqrt{...}$ | What's under the root |
Key test: After choosing $u$, check if $du$ (or a constant multiple of it) appears in the integral.
Find $\int 2x(x^2 + 1)^5 \, dx$.
Step 1: Choose u
The "inside" function is $x^2 + 1$.
Let $u = x^2 + 1$
Step 2: Find du
$$\frac{du}{dx} = 2x$$ $$du = 2x \, dx$$
Step 3: Substitute
$$\int 2x(x^2 + 1)^5 \, dx = \int u^5 \, du$$
Step 4: Integrate
$$\int u^5 \, du = \frac{u^6}{6} + C$$
Step 5: Back-substitute
$$= \frac{(x^2 + 1)^6}{6} + C$$
$$\boxed{\frac{(x^2 + 1)^6}{6} + C}$$
Find $\int x\sqrt{9 + x^2} \, dx$.
Step 1: Choose u
Let $u = 9 + x^2$
Step 2: Find du
$$du = 2x \, dx$$
We have $x \, dx$ in the integral, not $2x \, dx$. That's okay — solve for $x \, dx$:
$$x \, dx = \frac{1}{2} du$$
Step 3: Substitute
$$\int x\sqrt{9 + x^2} \, dx = \int \sqrt{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int u^{1/2} \, du$$
Step 4: Integrate
$$\frac{1}{2} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C$$
Step 5: Back-substitute
$$= \frac{1}{3}(9 + x^2)^{3/2} + C$$
$$\boxed{\frac{1}{3}(9 + x^2)^{3/2} + C}$$
Find $\int e^{3x} \, dx$.
Step 1: Choose u
Let $u = 3x$
Step 2: Find du
$$du = 3 \, dx \quad \Rightarrow \quad dx = \frac{1}{3} du$$
Step 3: Substitute
$$\int e^{3x} \, dx = \int e^u \cdot \frac{1}{3} \, du = \frac{1}{3} \int e^u \, du$$
Step 4: Integrate
$$\frac{1}{3} e^u + C$$
Step 5: Back-substitute
$$\boxed{\frac{1}{3} e^{3x} + C}$$
Find $\int \tan^2\theta \sec^2\theta \, d\theta$.
Step 1: Choose u
Since $\frac{d}{d\theta}(\tan\theta) = \sec^2\theta$, let:
$u = \tan\theta$
Step 2: Find du
$$du = \sec^2\theta \, d\theta$$
Step 3: Substitute
$$\int \tan^2\theta \sec^2\theta \, d\theta = \int u^2 \, du$$
Step 4: Integrate
$$\frac{u^3}{3} + C$$
Step 5: Back-substitute
$$\boxed{\frac{\tan^3\theta}{3} + C}$$
Find $\int \frac{x}{x^2 + 4} \, dx$.
Step 1: Choose u
Let $u = x^2 + 4$
Step 2: Find du
$$du = 2x \, dx \quad \Rightarrow \quad x \, dx = \frac{1}{2} du$$
Step 3: Substitute
$$\int \frac{x}{x^2 + 4} \, dx = \int \frac{1}{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int \frac{1}{u} \, du$$
Step 4: Integrate
$$\frac{1}{2} \ln|u| + C$$
Step 5: Back-substitute
$$\boxed{\frac{1}{2} \ln|x^2 + 4| + C}$$
(We can drop the absolute value since $x^2 + 4 > 0$ always.)
Find $\int \frac{\sin 2x}{1 + \cos^2 x} \, dx$.
Step 1: Simplify using identities
Recall $\sin 2x = 2\sin x \cos x$:
$$\int \frac{2\sin x \cos x}{1 + \cos^2 x} \, dx$$
Step 2: Choose u
Let $u = 1 + \cos^2 x$
Step 3: Find du
$$\frac{du}{dx} = 2\cos x \cdot (-\sin x) = -2\sin x \cos x$$ $$du = -2\sin x \cos x \, dx$$
So $2\sin x \cos x \, dx = -du$
Step 4: Substitute
$$\int \frac{2\sin x \cos x}{1 + \cos^2 x} \, dx = \int \frac{-du}{u} = -\int \frac{1}{u} \, du$$
Step 5: Integrate
$$-\ln|u| + C$$
Step 6: Back-substitute
$$\boxed{-\ln(1 + \cos^2 x) + C}$$
Evaluate $\int_0^2 x(x^2 + 1)^3 \, dx$.
Step 1: Choose u and find du
Let $u = x^2 + 1$, so $du = 2x \, dx$, meaning $x \, dx = \frac{1}{2} du$
Step 2: Change the limits!
When $x = 0$: $u = 0^2 + 1 = 1$ When $x = 2$: $u = 2^2 + 1 = 5$
Step 3: Substitute everything (including limits)
$$\int_0^2 x(x^2 + 1)^3 \, dx = \int_1^5 u^3 \cdot \frac{1}{2} \, du = \frac{1}{2} \int_1^5 u^3 \, du$$
Step 4: Integrate and evaluate
$$\frac{1}{2} \left[ \frac{u^4}{4} \right]_1^5 = \frac{1}{8} \left[ u^4 \right]_1^5 = \frac{1}{8}(625 - 1) = \frac{624}{8} = \boxed{78}$$
True or False: When using u-substitution, you can choose any part of the integrand as $u$.
False. The choice of $u$ must be strategic. You need $du$ (or a constant multiple of it) to appear in the remaining part of the integrand.
If you choose $u$ poorly, you'll end up with leftover $x$'s that you can't eliminate, and the substitution fails.
True or False: For definite integrals with u-substitution, you must always convert the limits of integration to $u$-values.
False. You have two valid options:
-
Change limits to $u$-values (recommended): Substitute the $x$-limits into $u = g(x)$ and evaluate directly in terms of $u$. No back-substitution needed.
-
Keep original limits: Do the indefinite integral, back-substitute to get $F(x)$, then evaluate $F(b) - F(a)$.
Both work, but changing the limits is usually cleaner and avoids back-substitution entirely.
True or False: If $\int f(g(x)) \cdot g'(x) \, dx$ can be solved by u-substitution with $u = g(x)$, the result is $F(g(x)) + C$ where $F$ is an antiderivative of $f$.
True. This is exactly what u-substitution does:
- Let $u = g(x)$, so $du = g'(x) \, dx$
- The integral becomes $\int f(u) \, du = F(u) + C$
- Back-substitute: $F(g(x)) + C$
This is the integration version of the chain rule.
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting to substitute ALL parts of the integral
Wrong: $\int 2x(x^2+1)^3 \, dx \to \int 2x \cdot u^3 \, dx$ (still has $x$ and $dx$!)
Why it's wrong: After substitution, the integral should only contain $u$ and $du$. If there are any $x$'s left, you haven't completed the substitution.
Correct: Since $u = x^2 + 1$ and $du = 2x \, dx$, replace $2x \, dx$ with $du$: $$\int 2x(x^2+1)^3 \, dx = \int u^3 \, du$$
❌ Mistake: Not changing limits in definite integrals
Wrong: $\int_0^1 2x(x^2+1)^3 \, dx = \int_0^1 u^3 \, du$ (limits still in terms of $x$!)
Why it's wrong: The limits 0 and 1 are $x$-values, but the integrand is now in terms of $u$. You must convert the limits.
Correct: When $x = 0$, $u = 1$. When $x = 1$, $u = 2$. $$\int_0^1 2x(x^2+1)^3 \, dx = \int_1^2 u^3 \, du$$
❌ Mistake: Choosing $u$ that leaves leftover $x$'s
Wrong: For $\int x^2 \sqrt{x+1} \, dx$, choosing $u = x + 1$ gives $du = dx$, leaving $x^2$ with no way to convert it.
Why it's wrong: You need to express everything in terms of $u$. Here you'd need $x^2 = (u-1)^2$, which works but creates more algebra.
Correct approach: Either use $x = u - 1$ to convert $x^2$, or consider if a different substitution might be cleaner.
U-Substitution (Reverse Chain Rule)
If you can identify an 'inside' function g(x) whose derivative g'(x) appears in the integrand, substitute u = g(x). The integral becomes ∫f(u) du, which is often much simpler.
Variables:
- $u$:
- the substitution variable (equals the 'inside' function g(x))
- $du$:
- the differential of u, equal to g'(x) dx
- $F$:
- an antiderivative of f
U-Substitution for Definite Integrals
When using u-substitution on a definite integral, convert the limits from x-values to u-values. Evaluate u = g(x) at each limit. No back-substitution needed!
Variables:
- $a, b$:
- original limits (in terms of x)
- $g(a), g(b)$:
- new limits (in terms of u)
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