Integration Using Substitution (College Board AP® Calculus BC): Study Guide

Determining indefinite integrals using u-substitutions

What is integration by substitution?

• Substitution simplifies an integral by defining an alternative variable (usually$u$) in terms of the original variable (usually$x$)

  • The integral in $u$ is much easier to solve than the original integral in $x$

  • The substitution can be reversed at the end to get the answer in terms of $x$

How do I integrate simple functions using u-substitution?

• In a simple integral involving substitution, you will usually be integrating a composite function (i.e., 'function of a function')

  • These can also be solved 'by inspection'

    • See the 'Integrals of Composite Functions' study guide

  • Substitution can be a safer method when 'by inspection' is awkward or difficult to spot

• STEP 1
  Identify the substitution to be used

  • It will be the secondary (or 'inside') function in a composite function

    • I.e. if the integral involves$f\left(g\right(x\left)\right)$, let$u=g\left(x\right)$

    • E.g. $\int 3x\cos \left(x^2-5\right)dx$

      • Let $u=x^2-5$

• STEP 2
  Differentiate the substitution and rearrange

  • $\frac{\mathrm{d}u}{\mathrm{d}x}$can be treated like a fraction (i.e. “multiply by$dx$" to get rid of fractions)

  • E.g. $u=x^2-5\Rightarrow \frac{du}{dx}=2x$

    • Then $du=2xdx\Rightarrow xdx=\frac{1}{2}du$

• STEP 3
  Replace all parts of the integral

  • All$x$ terms should be replaced with equivalent$u$ terms, including$dx$

  • E.g. $\int 3x\cos \left(x^2-5\right)dx=3\int \cos \left(x^2-5\right)·xdx$

    • So $\int 3x\cos \left(x^2-5\right)dx=3\int \cos u·\frac{1}{2}du=\frac{3}{2}\int \cos udu$

• STEP 4
  Integrate

  • E.g. $\frac{3}{2}\int \cos udu=\frac{3}{2}\sin u+C$

    • Don't forget the constant of integration

• STEP 5
  Substitute$x$ back in

  • Replace $u$ everywhere with the equivalent expression for $x$

  • E.g. $\frac{3}{2}\sin u+C=\frac{3}{2}\sin \left(x^2-5\right)+C$

    • So $\int 3x\cos \left(x^2-5\right)dx=\frac{3}{2}\sin \left(x^2-5\right)+C$

How do I integrate more complicated functions using u-substitution?

• The procedure here is exactly the same as for integrating simpler functions

  • However the substitution to use may not be as obvious

  • Practice questions like this to improve your integration by substitution skills

• E.g. $\int x\sqrt{x-4}dx$

  • Note that this is not an integral that can be solved 'by inspection' (i.e. by the 'reverse chain rule')

  • Identify the substitution

    • Let $u=x-4$

  • Differentiate the substitution and rearrange

    • $\frac{du}{dx}=1\Rightarrow du=dx$

  • Replace all parts of the integral

    • $u=x-4\Rightarrow x=u+4$

    • So $\int x\sqrt{x-4}dx=\int \left(u+4\right)\sqrt{u}du$

  • Integrate

    • $\int \left(u+4\right)\sqrt{u}du=\int \left(u+4\right)u^{\frac{1}{2}}du=\int \left(u^{\frac{3}{2}}+4u^{\frac{1}{2}}\right)du=\frac{2}{5}{u}^{\frac{5}{2}}+\frac{8}{3}{u}^{\frac{3}{2}}+C$

  • Substitute $x$ back in

    • $\int x\sqrt{x-4}dx=\frac{2}{5}{\left(x-4\right)}^{\frac{5}{2}}+\frac{8}{3}{\left(x-4\right)}^{\frac{3}{2}}+C$

Evaluating definite integrals using u-substitutions

How do I evaluate definite integrals using u-substitution?

• Definite integrals can also be solved using u-substitution

  • You just need to rewrite the integration limits in terms of u as well

• E.g. ${\int }_4^{8}x\sqrt{x-4}dx$

  • We've already seen that this can be integrated using the substitution $u=x-4$

    • with $du=dx$ and $x=u+4$ following from this

  • We just need to change the integral limits as well

    • When $x=4$, $u=4-4=0$

    • When $x=8$, $u=8-4=4$

  • Now when we replace all parts of the integral we get

    • ${\int }_4^{8}x\sqrt{x-4}dx={\int }_0^4\left(u+4\right)\sqrt{u}du$

  • Integrate, then evaluate the definite integral using the $u$ values

    • $\begin{array}{rcl}{\int }_0^4\left(u+4\right)\sqrt{u}du& =& {\left[\frac{2}{5}{u}^{\frac{5}{2}}+\frac{8}{3}{u}^{\frac{3}{2}}\right]}_0^4=\frac{512}{15}\end{array}$

  • Note that there is no need to substitute $x$ back in to evaluate the definite integral!