Integrals That Give tan⁻¹ and sin⁻¹

An integral that is an angle

$\frac{1}{{1 + x^{2}}}$ is not a power of x, and its numerator is not the derivative of its denominator, which would be 2x. Yet it is a derivative: the derivative of $\tan^{-1}x$.

Here is why. If $y = \tan^{-1}x$, then x = tan y, and $\frac{dx}{dy} = \sec^{2}y = 1 + \tan^{2}y = 1 + x^{2}$. So $\frac{dy}{dx} = \frac{1}{{1 + x^{2}}}$. Read backwards, $\int \frac{1}{{1 + x^{2}}} \,dx = \tan^{-1}x + C$.

An area of $\frac{\pi}{4}$

The area under $y = \frac{1}{{1 + x^{2}}}$ from x = 0 to x = 1 is $\tan^{-1}1 - \tan^{-1}0 = \frac{\pi}{4} - 0 = \frac{\pi}{4} \approx 0.785$. An area under a curve comes out as an angle: the angle whose tangent is 1.

From 0 to 10 the area is $\tan^{-1}10 \approx 1.471$. As the right-hand edge moves further out, the area keeps growing, but $\tan^{-1}x$ stays below $\frac{\pi}{2} \approx 1.571$, so the area never passes $\frac{\pi}{2}$.

The curve $y = \frac{1}{{1 + x^{2}}}$, from height 1 at x = 0 to height ½ at x = 1, shaded between them. The shaded area is $\tan^{-1}1 = \frac{\pi}{4} \approx 0.785$.

$\sin^{-1}$ in the same way

If $y = \sin^{-1}x$, then x = sin y, and $\frac{dx}{dy} = \cos y$. With y between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$, cos y is not negative, so $\cos y = \sqrt{1 - \sin^{2}y} = \sqrt{1 - x^{2}}$. So $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^{2}}}$, and $\int \frac{1}{\sqrt{1 - x^{2}}} \,dx = \sin^{-1}x + C$, for x between −1 and 1.

The area under $y = \frac{1}{\sqrt{1 - x^{2}}}$ from 0 to ½ is $\sin^{-1}\frac{1}{2} - \sin^{-1}0 = \frac{\pi}{6} \approx 0.524$.

The curve $y = \frac{1}{\sqrt{1 - x^{2}}}$, which has height 1 at x = 0 and climbs without limit toward x = −1 and x = 1. The region from 0 to ½ is shaded, and its area is $\sin^{-1}\frac{1}{2} = \frac{\pi}{6} \approx 0.524$.

With $a^{2}$ in place of 1

For $\frac{1}{{a^{2} + x^{2}}}$, with a > 0, substitute x = au, so dx = a du. The denominator becomes $a^{2} + a^{2}u^{2} = a^{2}\left(1 + u^{2}\right)$, and the integral becomes $\int \frac{a}{{a^{2}\left(1 + u^{2}\right)}} \,du = \left(\frac{1}{a}\right) \int \frac{1}{{1 + u^{2}}} \,du = \left(\frac{1}{a}\right) \tan^{-1}u + C$. Putting $u = \frac{x}{a}$ back: $\int \frac{1}{{a^{2} + x^{2}}} \,dx = \left(\frac{1}{a}\right) \tan^{-1}\left(\frac{x}{a}\right) + C$.

For $\frac{1}{{9 + x^{2}}}$, a = 3, since 9 = 3 × 3. So the integral is $\left(\frac{1}{3}\right) \tan^{-1}\left(\frac{x}{3}\right) + C$. Check: differentiating gives $\left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right) \times \frac{1}{{1 + \frac{x^{2}}{9}}} = \left(\frac{1}{9}\right) \times \frac{9}{{9 + x^{2}}} = \frac{1}{{9 + x^{2}}}$. From 0 to 3 the integral is $\left(\frac{1}{3}\right)\left(\frac{\pi}{4}\right) = \frac{\pi}{12} \approx 0.262$.

For the root, the a cancels: $\sqrt{a^{2} - a^{2}u^{2}} = a\sqrt{1 - u^{2}}$, and dx = a du divided by that leaves $\frac{du}{\sqrt{1 - u^{2}}}$. So $\int \frac{1}{\sqrt{a^{2} - x^{2}}} \,dx = \sin^{-1}\left(\frac{x}{a}\right) + C$, with no $\frac{1}{a}$ in front. From 0 to 1, $\int \frac{1}{\sqrt{4 - x^{2}}} \,dx = \sin^{-1}\frac{1}{2} = \frac{\pi}{6}$.

A number in front of $x^{2}$ is handled the same way. $1 + 4x^{2} = 1 + \left(2x\right)^{2}$, so $\int \frac{1}{{1 + 4x^{2}}} \,dx = \frac{1}{2} \tan^{-1}\left(2x\right) + C$, the ½ undoing the 2 the chain rule brings out. From 0 to ½ that is $\frac{1}{2} \tan^{-1}1 = \frac{\pi}{8} \approx 0.393$.

The sign decides

$x^{2} - 1$ factors as (x − 1)(x + 1), so $\frac{1}{{x^{2} - 1}}$ splits into partial fractions and integrates to logs: ½ ln|x − 1| − ½ ln|x + 1| + C. $x^{2} + 1$ is never 0, so it has no factors, and $\frac{1}{{x^{2} + 1}}$ integrates to $\tan^{-1}x + C$ instead.

The numerator matters too. $\frac{x}{{1 + x^{2}}}$ is half of $\frac{2x}{{1 + x^{2}}}$, whose numerator is the derivative of its denominator, so it integrates to a log: $\frac{1}{2} \ln\left(1 + x^{2}\right) + C$. Only a constant over $1 + x^{2}$ gives $\tan^{-1}$.

The usual mistakes

Expecting a log. $\ln\left(1 + x^{2}\right)$ differentiates to $\frac{2x}{{1 + x^{2}}}$, which has an x on top that $\frac{1}{{1 + x^{2}}}$ does not.

Mixing up the pair. $\tan^{-1}$ goes with $1 + x^{2}$ and no root; $\sin^{-1}$ goes with the root of $1 - x^{2}$.

Leaving out the $\frac{1}{a}$. $\tan^{-1}\left(\frac{x}{3}\right)$ differentiates to $\frac{3}{{9 + x^{2}}}$, three times too much.

Using $a^{2}$ for a. In $\frac{1}{{4 + x^{2}}}$, a = 2, so the answer is $\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C$, not $\frac{1}{4} \tan^{-1}\left(\frac{x}{4}\right) + C$.

A factor $\frac{1}{a}$ in front of $\sin^{-1}\left(\frac{x}{a}\right)$. The a in the root cancels the a in dx, so nothing is left over.