5.7 Integrals Resulting in Inverse Trigonometric Functions

Big idea

The derivatives of the inverse trigonometric functions are algebraic: no sines or cosines appear in them, only square roots and sums of squares. Read backwards, that is a striking fact. It means certain purely algebraic integrands, which look as though they should have algebraic antiderivatives, do not. Their antiderivatives are angles.

Two shapes account for nearly all of this. A square root of a constant minus a square in the denominator produces an arcsine. A constant plus a square in the denominator produces an arctangent. Everything else in the section is machinery for putting a given integrand into one of those two shapes, usually by a linear substitution or by completing the square.

The generalized versions carry a constant $a$, and the position of $a$ in the answer is worth learning deliberately rather than rediscovering. In the arcsine form, $a$ appears only inside, dividing the variable. In the arctangent form it appears twice: inside dividing the variable, and outside as a factor of $1/a$. That asymmetry is not arbitrary; it is what the chain rule forces if the derivative is to come back with no leftover constant.

The other half of the work is discrimination. Integrands that differ by a single factor of $x$ in the numerator go to completely different places: one to an inverse trigonometric function, the other to a logarithm or a root by an ordinary substitution. Reading the numerator before choosing a method is the habit that this section is really training.

Decoder

For a constant $a > 0$, $\int \frac{du}{\sqrt{a^2-u^2}} = \arcsin\frac{u}{a} + C$ and $\int \frac{du}{a^2+u^2} = \frac{1}{a}\arctan\frac{u}{a} + C$.

Check the second one to see where the outside factor comes from. Differentiating $\arctan(u/a)$ with respect to $u$ gives $\frac{1}{1+(u/a)^2}\cdot\frac{1}{a}$, and multiplying numerator and denominator by $a^2$ turns that into $\frac{a}{a^2+u^2}$. One factor of $a$ too many, so the formula divides it out. The same computation on $\arcsin(u/a)$ produces $\frac{1}{\sqrt{a^2-u^2}}$ with nothing left over, which is why no outside factor appears there.

The arcsine form comes with a domain condition that the arctangent form does not. It requires $|u| < a$, since otherwise the expression under the root is not positive. An interval of integration reaching past $u = \pm a$ is a signal that something is wrong with the setup.

The classic mistake is pattern-matching on the denominator alone. The integrands $\frac{1}{1+x^2}$ and $\frac{x}{1+x^2}$ share a denominator and share nothing else: the first gives $\arctan x$, and the second gives $\tfrac{1}{2}\ln(1+x^2)$ by substitution. When an $x$ appears in the numerator, try the substitution $u = $ denominator first, and reserve the inverse trigonometric forms for the cases where the numerator is constant.

Definitions and results

Arcsine form. For $a > 0$ and $|u| < a$,

$$ \int \frac{du}{\sqrt{a^2-u^2}} = \arcsin\frac{u}{a} + C $$

Arctangent form. For $a > 0$ and all real $u$,

$$ \int \frac{du}{a^2+u^2} = \frac{1}{a}\arctan\frac{u}{a} + C $$

Arcsecant form. For $a > 0$ and $|u| > a$,

$$ \int \frac{du}{u\sqrt{u^2-a^2}} = \frac{1}{a}\,\text{arcsec}\,\frac{|u|}{a} + C $$

Linear substitutions. A coefficient on the variable is handled by absorbing it into $u$. For $\int \frac{dx}{b^2 + c^2x^2}$, take $u = cx$, which contributes $du = c\,dx$ and leaves a factor of $1/c$ outside.

Completing the square. A quadratic with a linear term becomes one of the standard shapes after rewriting. For instance $x^2 - 6x + 13 = (x-3)^2 + 4$, which is the arctangent shape with $u = x-3$ and $a = 2$. Complete the square whenever the denominator is quadratic and does not factor nicely.

Numerator test. A constant numerator over a sum or difference of squares points to an inverse trigonometric answer. A numerator proportional to the derivative of the quadratic points to a logarithm or a root. If both appear, split the integral into two pieces and handle each with its own method.

Hidden inner functions. The variable in the pattern does not have to be $x$. Since $e^{2x} = (e^x)^2$, the substitution $u = e^x$ turns $\int \frac{e^{x}}{1+e^{2x}}\,dx$ into $\int \frac{du}{1+u^2} = \arctan(e^x) + C$. The same trick handles $\sin x$ and $\ln x$ in that position.

Definite integrals. Change the limits with the substitution, as with any change of variable, and read the final answer as a difference of two angles in radians.

Worked examples

An arcsine with a constant

Evaluate $\displaystyle\int_0^{2}\frac{dx}{\sqrt{16-x^2}}$.

The denominator has the arcsine shape with $a^2 = 16$, so $a = 4$, and $u = x$ needs no substitution. The interval $[0,2]$ sits inside $(-4,4)$, so the formula applies throughout:

$$ \int_0^{2}\frac{dx}{\sqrt{16-x^2}} = \left[\arcsin\frac{x}{4}\right]_0^{2} = \arcsin\frac{1}{2} - \arcsin 0 = \frac{\pi}{6} $$

Note that the answer is an angle even though nothing trigonometric appeared in the problem.

An arctangent with a coefficient

Find $\displaystyle\int\frac{dx}{4+9x^2}$, then evaluate it from $0$ to $\tfrac{2}{3}$.

Write $9x^2 = (3x)^2$ and set $u = 3x$, so $du = 3\,dx$ and $dx = \tfrac{1}{3}\,du$. With $a = 2$,

$$ \int\frac{dx}{4+9x^2} = \frac{1}{3}\int\frac{du}{4+u^2} = \frac{1}{3}\cdot\frac{1}{2}\arctan\frac{u}{2} + C = \frac{1}{6}\arctan\frac{3x}{2} + C $$

Check by differentiating: $\frac{1}{6}\cdot\frac{1}{1+(3x/2)^2}\cdot\frac{3}{2} = \frac{1}{4}\cdot\frac{1}{1 + 9x^2/4} = \frac{1}{4+9x^2}$.

For the definite version, the limits $x = 0$ and $x = \tfrac{2}{3}$ become $u = 0$ and $u = 2$, so the argument of the arctangent runs from $0$ to $1$:

$$ \int_0^{2/3}\frac{dx}{4+9x^2} = \frac{1}{6}\left(\arctan 1 - \arctan 0\right) = \frac{1}{6}\cdot\frac{\pi}{4} = \frac{\pi}{24} $$

Completing the square

Evaluate $\displaystyle\int_3^{5}\frac{dx}{x^2-6x+13}$.

The quadratic has no real roots, so it will not factor; complete the square instead. Half of $-6$ is $-3$, and $(x-3)^2 = x^2-6x+9$, leaving $13 - 9 = 4$:

$$ x^2-6x+13 = (x-3)^2 + 4 $$

Substituting $u = x-3$ gives $du = dx$ and moves the limits from $x = 3, 5$ to $u = 0, 2$. With $a = 2$,

$$ \int_3^5\frac{dx}{x^2-6x+13} = \int_0^2 \frac{du}{u^2+4} = \frac{1}{2}\left[\arctan\frac{u}{2}\right]_0^2 = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8} $$

A sanity check on size: the denominator ranges from $4$ to $8$ on this interval, so the integrand is between $\tfrac{1}{8}$ and $\tfrac{1}{4}$ across a width of $2$, forcing a value between $0.25$ and $0.5$. Since $\pi/8 \approx 0.393$, the answer is in range.

Two integrands that look alike

Evaluate $\displaystyle\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}}$ and $\displaystyle\int_0^{1/2}\frac{x\,dx}{\sqrt{1-x^2}}$.

The first is the arcsine form with $a = 1$:

$$ \int_0^{1/2}\frac{dx}{\sqrt{1-x^2}} = \arcsin\frac{1}{2} - \arcsin 0 = \frac{\pi}{6} $$

The second has an $x$ upstairs, which is the derivative of $1-x^2$ up to the constant $-2$. Set $u = 1-x^2$, so $du = -2x\,dx$ and the limits become $u = 1$ and $u = \tfrac{3}{4}$:

$$ \int_0^{1/2}\frac{x\,dx}{\sqrt{1-x^2}} = -\frac{1}{2}\int_{1}^{3/4}u^{-1/2}\,du = -\big[u^{1/2}\big]_{1}^{3/4} = 1 - \frac{\sqrt{3}}{2} $$

That is about $0.134$, while the first integral is about $0.524$. One factor of $x$ changed both the method and the kind of answer.

Practice

Start with the forms themselves: recognizing the arcsine, arctangent and arcsecant patterns, identifying $a$, and writing the antiderivative with the outside factor where it belongs.

Then the definite versions, where a substitution or a completed square comes first and the limits move with it.

Quiz

Five items on the inverse trigonometric antiderivative forms, completing the square, and choosing between these and an ordinary substitution.