1.7 Integrals Resulting in Inverse Trigonometric Functions

Integrals Resulting in Inverse Trigonometric Functions

Inverse trigonometric function integrals

Certain integrals produce inverse trig functions as their antiderivatives. Recognizing these forms quickly is the core skill here. Each formula has a constant $a > 0$ that controls the domain and appears in the argument of the result.

The three formulas you need to know:

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

These formulas come directly from differentiating the corresponding inverse trig functions. For example, since $\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$, the arcsine integral formula is just the antiderivative version of that derivative (scaled by $a$).

The key to using them: look at the structure of your integrand and match it to one of these three patterns. The constant $a$ is whatever value sits in the spot where $a$ appears in the formula. The argument of the resulting inverse trig function is always the ratio $\frac{x}{a}$.

Substitution for inverse trigonometric integrals

Many integrals won't match the standard forms exactly. You'll need to manipulate them first, usually through completing the square or $u$-substitution, and sometimes through trig substitution.

Completing the square + $u$-sub (most common technique):

When the integrand has a quadratic that doesn't obviously match $a^2 - x^2$ or $a^2 + x^2$, complete the square first.

Example: Evaluate $\int \frac{1}{x^2 + 6x + 13}\, dx$.

1. Complete the square in the denominator: $x^2 + 6x + 13 = (x+3)^2 + 4$
2. Substitute $u = x + 3$, so $du = dx$
3. The integral becomes $\int \frac{1}{u^2 + 4}\, du$
4. This matches the arctangent form with $a = 2$
5. Result: $\frac{1}{2}\tan^{-1}\!\left(\frac{u}{2}\right) + C = \frac{1}{2}\tan^{-1}\!\left(\frac{x+3}{2}\right) + C$

Trig substitution (for radicals):

When you have a square root that resists simpler methods, trig substitution transforms the integral into a trig form you can evaluate.

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

After substituting, simplify using identities like $1 - \sin^2\theta = \cos^2\theta$, evaluate the resulting trig integral, then substitute back to express the answer in terms of $x$.

A common mistake: forgetting to check whether a simple $u$-sub works before jumping to trig substitution. Always scan for a direct $u$-sub first. For instance, $\int \frac{x}{\sqrt{4 - x^2}}\, dx$ has an $x$ in the numerator, so $u = 4 - x^2$ works cleanly. No trig sub needed.

Domain and geometry of inverse trigonometric integrals

Each formula has domain restrictions that come from the integrand itself:

• Arcsine form ($\sqrt{a^2 - x^2}$ in denominator): you need $a^2 - x^2 > 0$, so $|x| < a$. The expression under the radical can't be zero or negative.
• Arctangent form ($a^2 + x^2$ in denominator): this is always positive, so $x$ can be any real number. No restrictions.
• Arcsecant form ($x\sqrt{x^2 - a^2}$ in denominator): you need $x^2 - a^2 > 0$, so $|x| > a$.

For definite integrals, make sure your limits of integration fall within the valid domain. If they don't, the integral doesn't exist (or needs to be treated as an improper integral).

Geometrically, $\int_0^t \frac{1}{\sqrt{a^2 - x^2}}\, dx = \sin^{-1}\!\left(\frac{t}{a}\right)$ gives the angle (in radians) of a sector on a circle of radius $a$. The arctangent integral similarly measures an angle, but one associated with the curve $y = \frac{1}{a^2 + x^2}$, which is a bell-shaped curve centered at the origin with height $\frac{1}{a^2}$.

Advanced techniques for inverse trigonometric integrals

• Integration by parts comes into play when the inverse trig function is itself part of the integrand (e.g., $\int \sin^{-1}(x)\, dx$). In that case, let $u = \sin^{-1}(x)$ and $dv = dx$.
• Partial fractions may be needed to break a complex rational function into simpler pieces, some of which match the arctangent form. For example, a denominator like $(x^2+1)(x+2)$ would be decomposed before integrating.
• Coefficient adjustments: if the coefficient of $x^2$ isn't 1, factor it out before matching to a standard form. For $\int \frac{1}{9x^2 + 4}\, dx$, rewrite the denominator as $9\!\left(x^2 + \frac{4}{9}\right)$, then apply the arctangent formula with $a = \frac{2}{3}$.