5.7 Integrals Resulting in Inverse Trigonometric Functions

Integrals Resulting in Inverse Trigonometric Functions

Inverse Trigonometric Function Integrals

Certain rational expressions and expressions involving square roots don't integrate into the usual polynomial, exponential, or logarithmic forms. Instead, they produce inverse trigonometric functions. Recognizing these patterns saves you from a lot of unnecessary work.

There are three standard forms you need to memorize:

• Arcsine: $\int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \sin^{-1}\!\left(\frac{x}{a}\right) + C$ for $|x| < |a|$
• Arctangent: $\int \frac{1}{a^2 + x^2}\, dx = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C$
• Arcsecant: $\int \frac{1}{x\sqrt{x^2 - a^2}}\, dx = \frac{1}{a}\sec^{-1}\!\left(\frac{|x|}{a}\right) + C$ for $|x| > |a|$

The way to identify which formula applies is to look at the denominator's structure:

• See $\sqrt{a^2 - x^2}$? Think arcsine.
• See $a^2 + x^2$ (no square root)? Think arctangent.
• See $x\sqrt{x^2 - a^2}$? Think arcsecant.

Once you match the form, plug in the value of $a$, evaluate, and don't forget $+C$.

Example: Evaluate $\int \frac{1}{\sqrt{9 - x^2}}\, dx$.

Here $a^2 = 9$, so $a = 3$. This matches the arcsine form directly:

$\int \frac{1}{\sqrt{9 - x^2}}\, dx = \sin^{-1}\!\left(\frac{x}{3}\right) + C$

Substitution for Inverse Trigonometric Forms

Many integrals won't match the standard forms right away. You'll need to manipulate them first, usually with a $u$-substitution or by completing the square.

Using $u$-substitution:

When the variable expression inside the integrand isn't just $x$, a simple $u$-sub can bring it into standard form.

Example: Evaluate $\int \frac{1}{4 + (x-3)^2}\, dx$.

1. Let $u = x - 3$, so $du = dx$.

2. The integral becomes $\int \frac{1}{4 + u^2}\, du$.

3. Now $a^2 = 4$, so $a = 2$. Apply the arctangent formula: $\frac{1}{2}\tan^{-1}\!\left(\frac{u}{2}\right) + C$

4. Substitute back: $\frac{1}{2}\tan^{-1}\!\left(\frac{x-3}{2}\right) + C$

Completing the square:

If you see something like $x^2 + 6x + 13$ in the denominator, complete the square to reveal the standard form: $x^2 + 6x + 13 = (x+3)^2 + 4$. Then use a $u$-sub with $u = x + 3$.

Trigonometric substitution is a more advanced technique where you replace $x$ itself with a trig expression. The three cases are:

• For $\sqrt{a^2 - x^2}$: substitute $x = a\sin\theta$, $dx = a\cos\theta\, d\theta$
• For $\sqrt{x^2 + a^2}$: substitute $x = a\tan\theta$, $dx = a\sec^2\theta\, d\theta$
• For $\sqrt{x^2 - a^2}$: substitute $x = a\sec\theta$, $dx = a\sec\theta\tan\theta\, d\theta$

After integrating in terms of $\theta$, you substitute back to express the answer in terms of $x$. For most problems in this section, though, a $u$-sub or completing the square will be enough. Trig substitution becomes more central in Calculus II.

Domains and Ranges in Integration

The inverse trig functions have restricted domains and ranges by definition. When your answer involves one of these functions, the argument has to fall within the allowed domain, or the answer doesn't make sense.

• For $\sin^{-1}\!\left(\frac{x}{a}\right)$, you need $|x| < |a|$ (the original integrand has $\sqrt{a^2 - x^2}$, which already enforces this).
• For $\tan^{-1}\!\left(\frac{x}{a}\right)$, there's no restriction on $x$. Any real number works.
• For $\sec^{-1}\!\left(\frac{|x|}{a}\right)$, you need $|x| > |a|$ (again, the square root $\sqrt{x^2 - a^2}$ enforces this).

These restrictions rarely cause trouble on their own because the integrand's domain naturally matches. But when you evaluate definite integrals, make sure your limits of integration stay within the valid domain.

Advanced Integration Techniques

Some problems layer inverse trig forms with other techniques. Here are the most common combinations you'll encounter:

• Partial fractions first: If the integrand is a complex rational function, decompose it. Some of the resulting terms may produce inverse trig integrals.
• Completing the square: As shown above, this is often the key step that reveals a hidden arctangent or arcsine form.
• Chain rule awareness: If the integral has the form $\int \frac{f'(x)}{\sqrt{a^2 - [f(x)]^2}}\, dx$, the $f'(x)$ in the numerator acts as the "$du$" from a substitution with $u = f(x)$, and the result is $\sin^{-1}\!\left(\frac{f(x)}{a}\right) + C$. Always check whether a function and its derivative are both present before reaching for heavier techniques.
• Coefficient adjustments: Sometimes you need to factor a constant out of the square root or denominator to match the standard form. For instance, $\int \frac{1}{\sqrt{5 - 3x^2}}\, dx$ requires rewriting as $\int \frac{1}{\sqrt{5}\,\sqrt{1 - \frac{3}{5}x^2}}\, dx$ and then substituting to match the arcsine pattern.