Tutorials with examples, detailed solutions, and exercises with answers on how to use the powerful technique of integration by substitution to find integrals.
Review Integration by Substitution
The method of integration by substitution may be used to easily compute complex integrals. Let us examine an integral of the form:
\[ \int_{a}^{b} f(g(x)) g'(x) \, dx \]Let us make the substitution \( u = g(x) \), hence \( \dfrac{du}{dx} = g'(x) \) and \( du = g'(x) \, dx \). With this substitution, the definite integral transforms to:
\[ \int_{a}^{b} f(g(x)) g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du \]In what follows, \( C \) represents the constant of integration added to indefinite integrals.
Examples and Their Solutions
Click on each example to view the step-by-step solution.
Example 1: Evaluate \( \displaystyle \int \sin(ax + b) \, dx \)
Let \( u = ax + b \), which gives \( \dfrac{du}{dx} = a \) or \( dx = \dfrac{1}{a} \, du \). The substitution helps compute the integral as follows:
\[ \begin{aligned} \int \sin(ax + b) \, dx &= \dfrac{1}{a} \int \sin(u) \, du \\ &= -\dfrac{1}{a} \cos(u) + C \\ &= -\dfrac{1}{a} \cos(ax + b) + C \end{aligned} \]Example 2: Evaluate \( \displaystyle \int e^{3x - 2} \, dx \)
Let \( u = 3x - 2 \), which gives \( \dfrac{du}{dx} = 3 \) or \( dx = \dfrac{1}{3} \, du \). Hence:
\[ \begin{aligned} \int e^{3x - 2} \, dx &= \int e^u \cdot \dfrac{1}{3} \, du \\ &= \dfrac{1}{3} e^u + C \\ &= \dfrac{1}{3} e^{3x - 2} + C \end{aligned} \]Example 3: Evaluate \( \displaystyle \int x (2x^2 + 5)^4 \, dx \)
Let \( u = 2x^2 + 5 \), which gives \( \dfrac{du}{dx} = 4x \), \( du = 4x \, dx \), or \( \dfrac{1}{4} du = x \, dx \). The substitution gives:
\[ \begin{aligned} \int x (2x^2 + 5)^4 \, dx &= \int \dfrac{1}{4} u^4 \, du \\ &= \dfrac{1}{20} u^5 + C \\ &= \dfrac{1}{20} (2x^2 + 5)^5 + C \end{aligned} \]Example 4: Evaluate \( \displaystyle \int x \sqrt{2x + 1} \, dx \)
Let \( u = 2x + 1 \), which gives \( \dfrac{du}{dx} = 2 \) and \( dx = \dfrac{1}{2} \, du \). Solve \( u = 2x + 1 \) for \( x \) to obtain \( x = \dfrac{1}{2}(u - 1) \). The substitution gives:
\[ \begin{aligned} \int x \sqrt{2x + 1} \, dx &= \int \dfrac{1}{2}(u - 1) \sqrt{u} \cdot \dfrac{1}{2} \, du \\ &= \dfrac{1}{4} \int (u - 1) u^{1/2} \, du \\ &= \dfrac{1}{4} \left( \dfrac{2}{5} u^{5/2} - \dfrac{2}{3} u^{3/2} \right) + C \\ &= \dfrac{(2x + 1)^{3/2} (3x - 1)}{15} + C \end{aligned} \]Example 5: Evaluate \( \displaystyle \int (x - 5)^{-4} \, dx \)
Let \( u = x - 5 \), which gives \( \dfrac{du}{dx} = 1 \). Substituting into the given integral:
\[ \begin{aligned} \int (x - 5)^{-4} \, dx &= \int u^{-4} \, du \\ &= -\dfrac{1}{3} u^{-3} + C \\ &= -\dfrac{1}{3} (x - 5)^{-3} + C \end{aligned} \]Example 6: Evaluate \( \displaystyle \int -x e^{x^2 + 2} \, dx \)
Let \( u = x^2 + 2 \), which gives \( \dfrac{du}{dx} = 2x \) and \( \dfrac{1}{2} du = x \, dx \). Substituting into the integral:
\[ \begin{aligned} \int -x e^{x^2 + 2} \, dx &= \int -e^u \cdot \dfrac{1}{2} \, du \\ &= -\dfrac{1}{2} \int e^u \, du \\ &= -\dfrac{1}{2} e^u + C \\ &= -\dfrac{1}{2} e^{x^2 + 2} + C \end{aligned} \]Example 7: Evaluate \( \displaystyle \int \cos(x) \sin^4(x) \, dx \)
Let \( u = \sin(x) \), which gives \( \dfrac{du}{dx} = \cos(x) \) or \( \cos(x) \, dx = du \). Substituting into the integral:
\[ \begin{aligned} \int \cos(x) \sin^4(x) \, dx &= \int u^4 \, du \\ &= \dfrac{1}{5} u^5 + C \\ &= \dfrac{1}{5} \sin^5(x) + C \end{aligned} \]Example 8: Evaluate \( \displaystyle \int \dfrac{3x}{4x + 1} \, dx \)
Let \( u = 4x + 1 \), which gives \( \dfrac{du}{dx} = 4 \) or \( dx = \dfrac{1}{4} \, du \). Solving for \( x \) gives \( x = \dfrac{1}{4}(u - 1) \). Substitute to obtain:
\[ \begin{aligned} \int \dfrac{3x}{4x + 1} \, dx &= \int 3 \cdot \dfrac{1}{4} \cdot \dfrac{u - 1}{u} \, du \\ &= \dfrac{3}{16} \int \left( 1 - \dfrac{1}{u} \right) du \\ &= \dfrac{3}{16} (u - \ln|u|) + C \\ &= \dfrac{3}{16} \bigl( 4x + 1 - \ln|4x + 1| \bigr) + C \end{aligned} \]Example 9: Evaluate \( \displaystyle \int \dfrac{x}{\sqrt{x - 2}} \, dx \)
Let \( u = x - 2 \), which gives \( \dfrac{du}{dx} = 1 \), \( dx = du \), and \( x = u + 2 \). Substitution yields:
\[ \begin{aligned} \int \dfrac{x}{\sqrt{x - 2}} \, dx &= \int \dfrac{u + 2}{\sqrt{u}} \, du \\ &= \int \left(u^{1/2} + 2u^{-1/2}\right) du \\ &= \dfrac{2}{3} u^{3/2} + 4\sqrt{u} + C \\ &= \dfrac{2}{3} (x - 2)^{3/2} + 4\sqrt{x - 2} + C \end{aligned} \]Example 10: Evaluate \( \displaystyle \int (x + 2)^3 (x + 4)^2 \, dx \)
Let \( u = x + 2 \), which gives \( \dfrac{du}{dx} = 1 \), \( dx = du \), and \( x = u - 2 \) (so \( x + 4 = u + 2 \)). Substituting this:
\[ \begin{aligned} \int (x + 2)^3 (x + 4)^2 \, dx &= \int u^3 (u + 2)^2 \, du \\ &= \int u^3 (u^2 + 4u + 4) \, du \\ &= \int (u^5 + 4u^4 + 4u^3) \, du \\ &= \dfrac{1}{6} u^6 + \dfrac{4}{5} u^5 + u^4 + C \\ &= \dfrac{1}{6} (x + 2)^6 + \dfrac{4}{5} (x + 2)^5 + (x + 2)^4 + C \end{aligned} \]Example 11: Evaluate \( \displaystyle \int \dfrac{2x + 3}{x^2 + 3x + 1} \, dx \)
Let \( u = x^2 + 3x + 1 \), which gives \( \dfrac{du}{dx} = 2x + 3 \) or \( (2x + 3) \, dx = du \). The substitution computes the integral as follows:
\[ \begin{aligned} \int \dfrac{2x + 3}{x^2 + 3x + 1} \, dx &= \int \dfrac{1}{u} \, du \\ &= \ln|u| + C \\ &= \ln|x^2 + 3x + 1| + C \end{aligned} \]Exercises
Use the table of integrals and the method of integration by substitution to find the integrals below:
- \( \displaystyle \int \cos(3x - 2) \, dx \)
- \( \displaystyle \int e^{4x - 7} \, dx \)
- \( \displaystyle \int x(4x^2 + 5)^4 \, dx \)
- \( \displaystyle \int \dfrac{1}{(x + 3)^3} \, dx \)
Answers to Above Exercises
- \( \dfrac{1}{3} \sin(3x - 2) + C \)
- \( \dfrac{1}{4} e^{4x - 7} + C \)
- \( \dfrac{1}{40} (4x^2 + 5)^5 + C \)
- \( -\dfrac{1}{2(x + 3)^2} + C \)