Integrating Using
Substitution
Master the u-substitution technique to reverse the chain rule and evaluate integrals of composite functions.
AP Calculus BC
Why Do We Need Substitution?
Not every integral can be solved with basic antiderivative rules. When the integrand is a composite function, a product of a function and its derivative, direct integration fails. Substitution lets us simplify the integrand by introducing a new variable.
The Chain Rule in Reverse
Recall the chain rule for differentiation:
$$\frac{d}{dx}\bigl[F(g(x))\bigr] = F'(g(x))\cdot g'(x)$$
Integrating both sides gives us the substitution pattern:
$$\int F'(g(x))\cdot g'(x)\,dx = F(g(x)) + C$$
Visualizing u-Substitution
The substitution cycle: transform → simplify and integrate → back-substitute.
The Substitution Rule
$$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$$
Indefinite vs. Definite Substitution
Indefinite Integral
1. Set \(u = g(x)\), find \(du = g'(x)\,dx\)
2. Rewrite in terms of \(u\)
3. Integrate
4. Back-substitute \(u = g(x)\)
5. Add \(+C\)
2. Rewrite in terms of \(u\)
3. Integrate
4. Back-substitute \(u = g(x)\)
5. Add \(+C\)
Definite Integral
1. Set \(u = g(x)\), find \(du = g'(x)\,dx\)
2. Change limits: \(x=a \to u=g(a)\), \(x=b \to u=g(b)\)
3. Integrate from \(g(a)\) to \(g(b)\)
4. Evaluate: no back-substitution needed
2. Change limits: \(x=a \to u=g(a)\), \(x=b \to u=g(b)\)
3. Integrate from \(g(a)\) to \(g(b)\)
4. Evaluate: no back-substitution needed
Choosing the Right \(u\)
The hardest part of substitution is picking the right \(u\). Here are guiding principles:
| Strategy | When to Use | Example |
|---|---|---|
| Inner function | Composite function visible | \(u = x^2\) in \(\int 2x\,e^{x^2}\,dx\) |
| Denominator | Fraction with derivative in numerator | \(u = 1+x^2\) in \(\int\frac{2x}{1+x^2}\,dx\) |
| Exponent | Exponential with non-linear exponent | \(u = 3x\) in \(\int e^{3x}\,dx\) |
| Argument of trig | Trig function of non-\(x\) argument | \(u = 5x\) in \(\int\cos(5x)\,dx\) |
Step-by-Step Method
1
Identify the inner function. Look for a composite structure \(f(g(x))\) and choose \(u = g(x)\).
2
Compute \(du\). Differentiate: \(du = g'(x)\,dx\). Solve for \(dx\) if needed.
3
Rewrite the integral. Replace all \(x\)-expressions with \(u\) and \(du\). No \(x\) should remain.
4
Integrate in \(u\). Apply standard antiderivative rules.
5
Back-substitute. Replace \(u\) with \(g(x)\) and add \(+C\) (indefinite) or evaluate at new limits (definite).
Example 1: Basic Substitution
$$\int 2x\cos(x^2)\,dx$$
1
Let \(u = x^2\). The inner function is \(x^2\).
2
Then \(du = 2x\,dx\). Notice \(2x\,dx\) is already present.
3
Rewrite: \(\displaystyle\int \cos(u)\,du\).
4
Integrate: \(\sin(u) + C\).
5
Back-substitute: \(\sin(x^2) + C\).
Example 2: Adjusting Constants
$$\int x^2\sqrt{x^3 + 1}\,dx$$
1
Let \(u = x^3 + 1\).
2
\(du = 3x^2\,dx\), so \(x^2\,dx = \tfrac{1}{3}\,du\).
3
Rewrite: \(\displaystyle\frac{1}{3}\int u^{1/2}\,du\).
4
Integrate: \(\displaystyle\frac{1}{3}\cdot\frac{2}{3}u^{3/2} + C = \frac{2}{9}u^{3/2} + C\).
5
Back-substitute: \(\dfrac{2}{9}(x^3+1)^{3/2} + C\).
Example 3: Definite Integral
$$\int_0^2 x\,e^{x^2}\,dx$$
1
Let \(u = x^2\), so \(du = 2x\,dx\) and \(x\,dx = \tfrac{1}{2}\,du\).
2
Change limits: when \(x=0\), \(u=0^2=0\); when \(x=2\), \(u=2^2=4\).
3
Rewrite: \(\displaystyle\frac{1}{2}\int_0^4 e^u\,du\).
4
Evaluate: \(\displaystyle\frac{1}{2}\bigl[e^u\bigr]_0^4 = \frac{1}{2}(e^4 - e^0) = \frac{e^4-1}{2}\).
Graphical Interpretation
The area under \(2x\cos(x^2)\) from \(x=0\) to \(x=\sqrt{\pi/2}\) equals the area under \(\cos(u)\) from \(u=0\) to \(u=\pi/2\). Substitution reshapes the curve but preserves the area.
Common Substitution Patterns
Power Rule
\(\int [g(x)]^n g'(x)\,dx\)
Let \(u = g(x)\)
Let \(u = g(x)\)
Exponential
\(\int e^{g(x)} g'(x)\,dx\)
Let \(u = g(x)\)
Let \(u = g(x)\)
Logarithmic
\(\int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)|+C\)
Let \(u = g(x)\)
Let \(u = g(x)\)
Example 4: Trig Inside Trig
$$\int \tan(x)\,dx = \int \frac{\sin(x)}{\cos(x)}\,dx$$
1
Rewrite \(\tan(x)\) as \(\frac{\sin(x)}{\cos(x)}\). This is a fraction where the derivative of the denominator appears (up to sign).
2
Let \(u = \cos(x)\), so \(du = -\sin(x)\,dx\) and \(\sin(x)\,dx = -du\).
3
Rewrite: \(\displaystyle\int\frac{-du}{u} = -\ln|u| + C\).
4
Back-substitute: \(-\ln|\cos(x)| + C = \ln|\sec(x)| + C\).
Common Mistakes to Avoid
✕ Forgetting to Change Limits
For definite integrals, if you switch to \(u\) but keep the original \(x\)-limits, the answer is wrong. Always compute \(u(a)\) and \(u(b)\).
✕ Leftover \(x\) Variables
After substituting, no \(x\) should remain. If it does, your choice of \(u\) may need rethinking, or you need to express \(x\) in terms of \(u\).
✕ Missing the Constant Factor
If \(du = 3x^2\,dx\) but the integral has \(x^2\,dx\), you must multiply and divide by 3. Ignoring this factor ruins the result.
Summary Comparison
| Integral Form | Substitution | Result |
|---|---|---|
| \(\int 2x\,e^{x^2}\,dx\) | \(u = x^2\) | \(e^{x^2}+C\) |
| \(\int \cos(3x)\,dx\) | \(u = 3x\) | \(\frac{1}{3}\sin(3x)+C\) |
| \(\int \frac{2x}{x^2+1}\,dx\) | \(u = x^2+1\) | \(\ln(x^2+1)+C\) |
| \(\int x\sqrt{1-x^2}\,dx\) | \(u = 1-x^2\) | \(-\frac{1}{3}(1-x^2)^{3/2}+C\) |
| \(\int \sec^2(x)\tan(x)\,dx\) | \(u = \tan(x)\) | \(\frac{1}{2}\tan^2(x)+C\) |
Question 1 of 5
Evaluate: \(\displaystyle\int 6x^2\cos(x^3)\,dx\)
Let \(u = x^3\), so \(du = 3x^2\,dx\) and \(x^2\,dx = \frac{1}{3}du\). The integral becomes \(6 \cdot \frac{1}{3}\int\cos(u)\,du = 2\sin(u)+C = 2\sin(x^3)+C\).
Question 2 of 5
Evaluate the definite integral: \(\displaystyle\int_0^1 \frac{2x}{(x^2+1)^2}\,dx\)
Let \(u = x^2+1\), \(du = 2x\,dx\). Limits: \(u(0)=1\), \(u(1)=2\). The integral becomes \(\int_1^2 u^{-2}\,du = \bigl[-u^{-1}\bigr]_1^2 = -\frac{1}{2}+1 = \frac{1}{2}\).
Question 3 of 5
Which substitution is best for \(\displaystyle\int e^{\sin x}\cos x\,dx\)?
The exponent \(\sin x\) is the inner function and its derivative \(\cos x\) is already present. Setting \(u = \sin x\) gives \(du = \cos x\,dx\), and the integral simplifies to \(\int e^u\,du = e^u + C = e^{\sin x}+C\).
Question 4 of 5
Evaluate: \(\displaystyle\int \frac{\ln x}{x}\,dx\)
Let \(u = \ln x\), so \(du = \frac{1}{x}\,dx\). The integral becomes \(\int u\,du = \frac{u^2}{2}+C = \frac{(\ln x)^2}{2}+C\). This is the power rule applied after substitution.
Question 5 of 5
A student evaluates \(\displaystyle\int_0^{\pi} \sin(x)\cos(x)\,dx\) using \(u=\sin x\) and gets \(0\). They then try \(u=\cos x\) and also get \(0\). What should they conclude?
Both substitutions are correct. With \(u=\sin x\): limits become \(0\to 0\), giving \(\int_0^0 u\,du=0\). With \(u=\cos x\): limits become \(1\to -1\), giving \(-\int_1^{-1} u\,du = -\bigl[\frac{u^2}{2}\bigr]_1^{-1} = -(\frac{1}{2}-\frac{1}{2})=0\). The function \(\sin x\cos x = \frac{1}{2}\sin(2x)\) is symmetric on \([0,\pi]\), so the integral is \(0\).