Integration by Substitution Notes

Integration by Substitution

Change of Variable of Integration

• Goal: Simplify the integral into a familiar form using a table of known integrals.
• Example: Indefinite integral of $(1 - 2x)^2$.
  • Let $u = 1 - 2x$.
  • Find $du$ in terms of $dx$:
    • $\frac{du}{dx} = -2$
    • $du = -2 \, dx$
    • $dx = \frac{du}{-2}$
  • Substitute $u$ and $dx$ into the integral:
    • $I = \int (1 - 2x)^2 \, dx = \int u^2 \, \frac{du}{-2} = -\frac{1}{2} \int u^2 \, du$
  • Integrate with respect to $u$:
    • $- \frac{1}{2} \int u^2 du = -\frac{1}{2} \cdot \frac{u^3}{3} + C = -\frac{1}{6} u^3 + C$
  • Substitute back $u = 1 - 2x$:
    • $I = -\frac{1}{6} (1 - 2x)^3 + C$
• Key Points:
  • Express the original integral in terms of the new variable $u$.
  • Ensure all $x$ terms are replaced with $u$ terms.
  • If $x$ terms remain, re-evaluate the substitution.

Definite Integrals

• Method 1: Use the antiderivative found via substitution and apply the Fundamental Theorem of Calculus using the original limits of integration.
• Method 2: Change the limits of integration to correspond to the new variable $u$.
  • If the original limits are $x = a$ to $x = b$, find the corresponding $u$ limits $u(a)$ and $u(b)$.
  • Integrate with respect to $u$ using the new limits.
• Example: Definite integral of $(1 - 2x)^2$ from $x = 0$ to $x = 1$.
  • Using the antiderivative: $-\frac{1}{6}(1 - 2x)^3 + C$
    • Evaluate at $x = 1$: $-\frac{1}{6}(1 - 2(1))^3 + C = \frac{1}{6} + C$
    • Evaluate at $x = 0$: $-\frac{1}{6}(1 - 2(0))^3 + C = -\frac{1}{6} + C$
    • Subtract: $(\frac{1}{6} + C) - (-\frac{1}{6} + C) = \frac{1}{3}$
  • Changing the limits:
    • Let $u = 1 - 2x$.
    • $du = -2 \, dx$
    • When $x = 0$, $u = 1 - 2(0) = 1$
    • When $x = 1$, $u = 1 - 2(1) = -1$
    • $I = \int<em>{0}^{1} (1 - 2x)^2 \, dx = \int</em>{1}^{-1} u^2 \, \frac{du}{-2} = -\frac{1}{2} \int_{1}^{-1} u^2 \, du$
    • Swap the limits and change the sign: $I = \frac{1}{2} \int_{-1}^{1} u^2 \, du$
    • Integrate: $I = \frac{1}{2} \cdot \frac{u^3}{3} \Big|<em>{-1}^{1} = \frac{1}{6} [u^3]</em>{-1}^{1} = \frac{1}{6} [(1)^3 - (-1)^3] = \frac{1}{6} [1 - (-1)] = \frac{1}{6} (2) = \frac{1}{3}$
      Note: When changing variables in definite integrals, you must either revert back to the original variable $x$ with the original limits, or convert the limits to the new variable $u$.

Reversing the Chain Rule

• Chain Rule: If $f$ and $g$ are differentiable, then $\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)$.
• Analogy: Peeling an orange, layer by layer.
• Example: Derivative of $\sin(\cos(x))$.
  • Let $f(x) = \sin(x)$ and $g(x) = \cos(x)$.
  • $\frac{d}{dx} [\sin(\cos(x))] = \cos(\cos(x)) \cdot (-\sin(x)) = -\sin(x) \cos(\cos(x))$
• Reversing the process:
  • If $\frac{d}{dx} [\sin(\cos(x))] = -\sin(x) \cos(\cos(x))$, then $\int -\sin(x) \cos(\cos(x)) \, dx = \sin(\cos(x)) + C$
• General Formula: $\int f'(g(x)) \cdot g'(x) \, dx = f(g(x)) + C$

Method of Substitution (Reversing the Chain Rule)

• Combine the change of variable with reversing the chain rule.
• Inner Function: $g(x)$ (the function inside another function).
• Outer Function: $f'(x)$ (the derivative of the outer function).
• Steps:
  • Let $u = g(x)$ (inner function).
  • Find $\frac{du}{dx} = g'(x)$.
  • Solve for $dx = \frac{du}{g'(x)}$.
  • Substitute $u$ and $dx$ into the integral.
  • Simplify and integrate with respect to $u$.
  • Substitute back $u = g(x)$ to get the final answer in terms of $x$.
• General Form: $\int f'(g(x)) \cdot g'(x) \, dx = \int f'(u) \, du = f(u) + C = f(g(x)) + C$
• Example: $\int 2x \cos(x^2 - 7) \, dx$
  • Let $u = x^2 - 7$
  • $\frac{du}{dx} = 2x$
  • $dx = \frac{du}{2x}$
  • $\int 2x \cos(u) \frac{du}{2x} = \int \cos(u) \, du = \sin(u) + C = \sin(x^2 - 7) + C$
• Example: $\int x \sqrt{x^2 + 1} \, dx$
  • Let $u = x^2 + 1$
  • $\frac{du}{dx} = 2x$
  • $dx = \frac{du}{2x}$
  • $\int x \sqrt{u} \frac{du}{2x} = \frac{1}{2} \int u^{\frac{1}{2}} \, du = \frac{1}{2} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{1}{3} u^{\frac{3}{2}} + C = \frac{1}{3} (x^2 + 1)^{\frac{3}{2}} + C$

Exercise

• Evaluate the definite integral:
  $\int_{1}^{e} \frac{(\ln x)^3}{x} \, dx$