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This cheat sheet covers how to use uu-substitution to evaluate integrals by reversing the chain rule. Students need it because many integrals look complicated until the inside function and its derivative are matched. The examples help students choose uu, rewrite dxdx, change bounds when needed, and avoid common algebra errors. It is designed as a formula-forward reference for grades 11-12 calculus practice. The core idea is to set u=g(x)u = g(x) when an integral contains a composite function such as f(g(x))g(x)f(g(x))g'(x). Then compute du=g(x)dxdu = g'(x)\,dx and rewrite the whole integral in terms of uu. For indefinite integrals, integrate in uu and substitute back to xx. For definite integrals, either change the bounds to uu-values or substitute back before applying the original xx-bounds.

Key Facts

  • The basic uu-substitution pattern is f(g(x))g(x)dx=f(u)du\int f(g(x))g'(x)\,dx = \int f(u)\,du when u=g(x)u = g(x) and du=g(x)dxdu = g'(x)\,dx.
  • For an indefinite integral, after finding f(u)du\int f(u)\,du, substitute u=g(x)u = g(x) back into the answer and add +C+C.
  • For a definite integral, if u=g(x)u = g(x), then abf(g(x))g(x)dx=g(a)g(b)f(u)du\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.
  • A constant multiplier can be adjusted because f(g(x))g(x)dx\int f(g(x))g'(x)\,dx may require multiplying and dividing by the same constant.
  • For powers, the pattern (g(x))ng(x)dx=(g(x))n+1n+1+C\int (g(x))^n g'(x)\,dx = \frac{(g(x))^{n+1}}{n+1}+C works when n1n \ne -1.
  • For logarithmic forms, g(x)g(x)dx=lng(x)+C\int \frac{g'(x)}{g(x)}\,dx = \ln|g(x)|+C.
  • For exponential forms, eg(x)g(x)dx=eg(x)+C\int e^{g(x)}g'(x)\,dx = e^{g(x)}+C.
  • For trigonometric forms, examples include cos(g(x))g(x)dx=sin(g(x))+C\int \cos(g(x))g'(x)\,dx = \sin(g(x))+C and sin(g(x))g(x)dx=cos(g(x))+C\int \sin(g(x))g'(x)\,dx = -\cos(g(x))+C.

Vocabulary

uu-substitution
A method for rewriting an integral using u=g(x)u = g(x) so the integral becomes simpler to evaluate.
Composite function
A function inside another function, such as sin(3x2)\sin(3x^2) where 3x23x^2 is the inside function.
Differential
The expression du=g(x)dxdu = g'(x)\,dx that tells how a small change in uu relates to a small change in xx.
Antiderivative
A function F(x)F(x) whose derivative is the integrand, so F(x)=f(x)F'(x) = f(x).
Definite integral bounds
The lower and upper limits of integration, which must be changed from xx-values to uu-values when using uu-substitution directly.
Constant of integration
The +C+C added to an indefinite integral because antiderivatives differ by a constant.

Common Mistakes to Avoid

  • Choosing only the outside function for uu is wrong because uu should usually be the inside expression g(x)g(x) whose derivative also appears in the integrand.
  • Forgetting to rewrite dxdx using dudu is wrong because the integral must be completely converted into the new variable before integrating.
  • Dropping a constant factor is wrong because if du=6xdxdu = 6x\,dx but the integral has 2xdx2x\,dx, the substitution needs the factor 13\frac{1}{3}.
  • Using the original bounds after changing to uu is wrong because bounds like x=ax = a and x=bx = b must become u=g(a)u = g(a) and u=g(b)u = g(b).
  • Forgetting to substitute back in an indefinite integral is wrong because the final antiderivative must be written in terms of the original variable unless the problem says otherwise.

Practice Questions

  1. 1 Evaluate 6x(3x2+5)4dx\int 6x(3x^2+5)^4\,dx using uu-substitution.
  2. 2 Evaluate 2xx2+9dx\int \frac{2x}{x^2+9}\,dx using uu-substitution.
  3. 3 Evaluate 014xe2x2dx\int_0^1 4x e^{2x^2}\,dx by changing the bounds to uu-values.
  4. 4 Explain why u=x2+1u = x^2+1 is a better substitution than u=x2+1u = \sqrt{x^2+1} for an integral containing xx2+1dx\frac{x}{\sqrt{x^2+1}}\,dx.