Integrating Expressions Involving [latex]\sqrt{{a}^{2}+{x}^{2}}[/latex]

Now let’s tackle expressions involving [latex]\sqrt{{a}^{2}+{x}^{2}}[/latex]. This requires a different approach than what we just learned.

The key difference here is domain. Unlike [latex]\sqrt{{a}^{2}-{x}^{2}}[/latex], the expression [latex]\sqrt{{a}^{2}+{x}^{2}}[/latex] is defined for all real values of [latex]x[/latex]. This means we need a trigonometric function that also has a range of all real numbers.

Our options:

• [latex]x = a\tan\theta[/latex] ✓ (tangent has range [latex](-\infty, \infty)[/latex])
• [latex]x = a\cot\theta[/latex] ✓ (cotangent has range [latex](-\infty, \infty)[/latex])

Either of these substitutions would actually work, but the standard substitution is [latex]x=a\tan\theta[/latex] or, equivalently, [latex]\tan\theta =\frac{x}{a}[/latex], where [latex]-\frac{\pi}{2} < \theta < \frac{\pi}{2}[/latex].

Why This Substitution Works

When you substitute [latex]x = a\tan\theta[/latex]:

[latex]\sqrt{{a}^{2}+{x}^{2}} = \sqrt{{a}^{2}+{(a\tan\theta)}^{2}} = \sqrt{{a}^{2}(1+{\tan}^{2}\theta)} = \sqrt{{a}^{2}{\sec}^{2}\theta} = a\sec\theta[/latex]

Since [latex]\tan\theta = \frac{x}{a}[/latex], you can build this reference triangle:

• Adjacent side: [latex]a[/latex]
• Opposite side: [latex]x[/latex]
• Hypotenuse: [latex]\sqrt{a^2+x^2}[/latex] (by Pythagorean theorem)
• Therefore: [latex]\theta = \tan^{-1}\left(\frac{x}{a}\right)[/latex]

The reference triangle assumes [latex]x > 0[/latex], but don’t worry! The trigonometric ratios it produces work for all values of [latex]x[/latex], including when [latex]x \leq 0[/latex].

Evaluate [latex]\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}[/latex] and check the solution by differentiating.

Show Solution

Begin with the substitution [latex]x=\tan\theta[/latex] and [latex]dx={\sec}^{2}\theta d\theta[/latex]. Since [latex]\tan\theta =x[/latex], draw the reference triangle in the following figure.

Thus,

[latex]\begin{array}{ccccc}\hfill {\displaystyle\int \frac{dx}{\sqrt{1+{x}^{2}}}}& ={\displaystyle\int \frac{{\sec}^{2}\theta }{\sec\theta }d\theta }\hfill & & & \begin{array}{c}\text{Substitute}x=\tan\theta \text{and}dx={\sec}^{2}\theta d\theta .\text{This}\hfill \\ \text{substitution makes}\sqrt{1+{x}^{2}}=\sec\theta .\text{Simplify.}\hfill \end{array}\hfill \\ & ={\displaystyle\int ^{\text{ }}\sec\theta d\theta }\hfill & & & \text{Evaluate the integral.}\hfill \\ & =\text{ln}|\sec\theta +\tan\theta |+C\hfill & & & \begin{array}{c}\text{Use the reference triangle to express the result}\hfill \\ \text{in terms of}x.\hfill \end{array}\hfill \\ & =\text{ln}|\sqrt{1+{x}^{2}}+x|+C.\hfill & & & \end{array}[/latex]

 

To check the solution, differentiate:

[latex]\begin{array}{cc}\hfill \frac{d}{dx}\left(\text{ln}|\sqrt{1+{x}^{2}}+x|\right)& =\frac{1}{\sqrt{1+{x}^{2}}+x}\cdot \left(\frac{x}{\sqrt{1+{x}^{2}}}+1\right)\hfill \\ & =\frac{1}{\sqrt{1+{x}^{2}}+x}\cdot \frac{x+\sqrt{1+{x}^{2}}}{\sqrt{1+{x}^{2}}}\hfill \\ & =\frac{1}{\sqrt{1+{x}^{2}}}.\hfill \end{array}[/latex]

 

Since [latex]\sqrt{1+{x}^{2}}+x>0[/latex] for all values of [latex]x[/latex], we could rewrite [latex]\text{ln}|\sqrt{1+{x}^{2}}+x|+C=\text{ln}\left(\sqrt{1+{x}^{2}}+x\right)+C[/latex], if desired.

In the example below, we explore how to use hyperbolic trigonometric functions as an alternative substitution method. First, we briefly review the definition of these functions and their derivatives.