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Integration by Substitution

Integration by Substitution

Integration by substitution, often called u-substitution, is a powerful technique in calculus used to evaluate integrals. You can think of it as the reverse of the chain rule for differentiation.

When you have an integral where the integrand contains a composite function and the derivative of the inner function is also present, substitution is the perfect tool.

The Method of U-Substitution

The general formula for substitution is based on the relationship: f(g(x))g(x)dx=f(u)du\int f(g(x))g'(x) \, dx = \int f(u) \, du

Here are the standard steps to solve an integral using substitution:

  1. Choose an inner function u=g(x)u = g(x). A good rule of thumb is to look for a part of the integrand whose derivative is also multiplying the rest of the expression.
  2. Find the differential du=g(x)dxdu = g'(x) \, dx.
  3. Substitute uu and dudu into the original integral to replace all instances of xx and dxdx.
  4. Evaluate the new integral with respect to uu.
  5. Replace uu with the original expression g(x)g(x) (for indefinite integrals).

Example 1: Indefinite Integral

Find 2xcos(x2)dx\int 2x \cos(x^2) \, dx

  1. Identify the inner function. Let u=x2u = x^2.
  2. Find the derivative: du=2xdxdu = 2x \, dx. Notice that 2xdx2x \, dx is exactly what we have in the original integral!
  3. Substitute uu and dudu into the integral: cos(u)du\int \cos(u) \, du
  4. Evaluate the integral: cos(u)du=sin(u)+C\int \cos(u) \, du = \sin(u) + C
  5. Substitute x2x^2 back in for uu: sin(x2)+C\sin(x^2) + C

Example 2: Definite Integral

For definite integrals, you must also change the limits of integration to match your new variable uu.

Evaluate 01x2(x3+1)4dx\int_0^1 x^2(x^3 + 1)^4 \, dx

  1. Let u=x3+1u = x^3 + 1.
  2. Find the derivative: du=3x2dxdu = 3x^2 \, dx. Our integral only has x2dxx^2 \, dx, so we can rewrite this as 13du=x2dx\frac{1}{3} \, du = x^2 \, dx.
  3. Change the limits of integration:
    • Lower limit: When x=0x = 0, u=(0)3+1=1u = (0)^3 + 1 = 1.
    • Upper limit: When x=1x = 1, u=(1)3+1=2u = (1)^3 + 1 = 2.
  4. Substitute everything into the integral: 12(u)4(13du)=1312u4du\int_1^2 (u)^4 \left(\frac{1}{3} \, du\right) = \frac{1}{3} \int_1^2 u^4 \, du
  5. Evaluate the definite integral: 13[u55]12=115(2515)=115(321)=3115\frac{1}{3} \left[ \frac{u^5}{5} \right]_1^2 = \frac{1}{15} (2^5 - 1^5) = \frac{1}{15} (32 - 1) = \frac{31}{15}