We choose a substitution that simplifies the integral.

If we let \(u = f(x)\), then we must also find: \[ \frac{du}{dx} \quad \Rightarrow \quad du = f'(x)\,dx \] Replace everything in the integral with \(u\), integrate, and then convert back to \(x\).

• You see a function AND its derivative nearby, e.g. \( (3x^2 + 1)^5 \cdot 6x \)
• Expressions involving composite functions, e.g. \( e^{2x+3} \)
• Square roots with linear expressions, e.g. \( \sqrt{5x - 1} \)
• Trig functions with inner angles, e.g. \( \sin(3x) \), \( \cos(5x) \)

Tip: The best substitution simplifies the inside of brackets or roots.

Example 1: \(\displaystyle \int (3x^2 + 1)^4 \cdot 6x \, dx\)

Let \(u = 3x^2 + 1\) → \(du = 6x\,dx\). \[ \int u^4 \, du = \frac{u^5}{5} + C = \frac{(3x^2 + 1)^5}{5} + C \]

Example 2: \(\displaystyle \int x \sqrt{x^2 + 4}\, dx\)

Let \(u = x^2 + 4\) → \(du = 2x\,dx\) → \(x\,dx = \frac{du}{2}\). \[ \int x\sqrt{x^2 + 4}\,dx = \int \frac{1}{2}\sqrt{u}\,du = \frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C = \frac{(x^2 + 4)^{3/2}}{3} + C \]

Example 3: \(\displaystyle \int e^{2x+3}\, dx\)

Let \(u = 2x + 3\) → \(du = 2dx\) → \(dx = \frac{du}{2}\). \[ \int e^{2x+3} dx = \frac{1}{2} \int e^u du = \frac{1}{2}e^u + C = \frac{1}{2}e^{2x+3} + C \]

Example 4 (Trig): \(\displaystyle \int \sin(4x)\,dx\)

Let \(u = 4x\) → \(du = 4 dx\) → \(dx = \frac{du}{4}\). \[ \int \sin(4x)\,dx = \frac{1}{4}\int \sin u\,du = -\frac{1}{4}\cos u + C = -\frac{1}{4}\cos(4x) + C \]

When using substitution for a definite integral: change the limits to the new variable.

Example: \(\displaystyle \int_{0}^{1} (2x + 1)^3 \cdot 2\,dx\)

Let \(u = 2x + 1\). Then:

• when \(x = 0\), \(u = 1\)
• when \(x = 1\), \(u = 3\)

Then: \[ \int_{0}^{1} (2x+1)^3 \cdot 2 dx = \int_{1}^{3} u^3 \, du = \left[\frac{u^4}{4}\right]_{1}^{3} = \frac{3^4 - 1^4}{4} = \frac{80}{4} = 20 \]

• Always choose the expression inside brackets, roots, or exponents as \(u\).
• If \(du\) does not appear, check algebra—usually a factor is missing.
• Factor out constants before substituting (e.g. \(\frac{1}{4}\)).
• For definite integrals, change the limits after substitution.
• Cambridge gives marks for correct substitution even if integration is wrong.