Find integrals efficiently using an integral table
Use technology to solve integration problems
Tables of Integrals
You’ve learned several integration methods, but there are additional tools that can help you tackle more complex integrals or verify your answers. One of the most useful resources is integration tables.
integration tables
Pre-computed lists of integrals and their antiderivatives that you can reference to evaluate or check your work quickly. You’ll find these in many calculus textbooks, including the appendices of this one.
Integration tables can be incredibly helpful, but you need to use them wisely. They’re great for:
Quick evaluation of integrals that match standard forms
Checking your work after solving an integral manually
Finding patterns that might help with similar problems
Keep in mind that two completely correct solutions can look very different, so don’t panic if your answer doesn’t match the table exactly—they might still be equivalent. Here’s a perfect example of how the same integral can have multiple correct forms:
Using trigonometric substitution with [latex]x=\tan\theta[/latex]:
These look completely different, but they’re actually the same! We can prove algebraically that [latex]{\text{sinh}}^{-1}x=\text{ln}\left(x+\sqrt{{x}^{2}+1}\right)[/latex].
Two antiderivatives are equivalent if their difference is just a constant. This makes sense because the derivative of a constant is zero, so [latex]\frac{d}{dx}[F(x) + C_1] = \frac{d}{dx}[G(x) + C_2][/latex] when [latex]F(x) - G(x) = \text{constant}[/latex].
to evaluate [latex]\displaystyle\int \frac{\sqrt{16-{e}^{2x}}}{{e}^{x}}dx[/latex].
If we look at integration tables, we see that several formulas contain expressions of the form [latex]\sqrt{{a}^{2}-{u}^{2}}[/latex]. This expression is actually similar to [latex]\sqrt{16-{e}^{2x}}[/latex], where [latex]a=4[/latex] and [latex]u={e}^{x}[/latex]. Keep in mind that we must also have [latex]du={e}^{x}[/latex]. Multiplying the numerator and the denominator of the given integral by [latex]{e}^{x}[/latex] should help to put this integral in a useful form. Thus, we now have
Substituting [latex]u={e}^{x}[/latex] and [latex]du={e}^{x}[/latex] produces [latex]\displaystyle\int \frac{\sqrt{{a}^{2}-{u}^{2}}}{{u}^{2}}du[/latex]. From the integration table (#88 in Appendix A),