Integrals That Give tan⁻¹ and sin⁻¹

An integral that is an angle

$\frac{1}{1+x^2}$1/(1 + x²) 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 ${\tan }^{-1}x$tan⁻¹x.

Here is why. If $y={\tan }^{-1}x$y = tan⁻¹x, then x = tan y, and $\frac{dx}{dy}={\sec }^{2}y=1+{\tan }^{2}y=1+x^2$dx/dy = sec²y = 1 + tan²y = 1 + x². So $\frac{dy}{dx}=\frac{1}{1+x^2}$dy/dx = 1/(1 + x²). Read backwards, $\int \frac{1}{1+x^2}dx={\tan }^{-1}x+C$∫ 1/(1 + x²) dx = tan⁻¹x + C.

An area of $\frac{\pi }{4}$π/4

The area under $y=\frac{1}{1+x^2}$y = 1/(1 + x²) from x = 0 to x = 1 is ${\tan }^{-1}1-{\tan }^{-1}0=\frac{\pi }{4}-0=\frac{\pi }{4}\approx 0.785$tan⁻¹1 − tan⁻¹0 = π/4 − 0 = π/4 ≈ 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 ${\tan }^{-1}10\approx 1.471$tan⁻¹10 ≈ 1.471. As the right-hand edge moves further out, the area keeps growing, but ${\tan }^{-1}x$tan⁻¹x stays below $\frac{\pi }{2}\approx 1.571$π/2 ≈ 1.571, so the area never passes $\frac{\pi }{2}$π/2.

The curve $y=\frac{1}{1+x^2}$y = 1/(1 + x²), from height 1 at x = 0 to height ½ at x = 1, shaded between them. The shaded area is ${\tan }^{-1}1=\frac{\pi }{4}\approx 0.785$tan⁻¹1 = π/4 ≈ 0.785.

${\sin }^{-1}$sin⁻¹ in the same way

If $y={\sin }^{-1}x$y = sin⁻¹x, then x = sin y, and $\frac{dx}{dy}=\cos y$dx/dy = cos y. With y between $-\frac{\pi }{2}$−π/2 and $\frac{\pi }{2}$π/2, cos y is not negative, so $\cos y=\sqrt{1-{\sin }^{2}y}=\sqrt{1-x^2}$cos y = √(1 − sin²y) = √(1 − x²). So $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$dy/dx = 1/√(1 − x²), and $\int \frac{1}{\sqrt{1-x^2}}dx={\sin }^{-1}x+C$∫ 1/√(1 − x²) dx = sin⁻¹x + C, for x between −1 and 1.

The area under $y=\frac{1}{\sqrt{1-x^2}}$y = 1/√(1 − x²) from 0 to ½ is ${\sin }^{-1}\frac{1}{2}-{\sin }^{-1}0=\frac{\pi }{6}\approx 0.524$sin⁻¹½ − sin⁻¹0 = π/6 ≈ 0.524.

The curve $y=\frac{1}{\sqrt{1-x^2}}$y = 1/√(1 − x²), 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 ${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}\approx 0.524$sin⁻¹½ = π/6 ≈ 0.524.

With $a^2$a² in place of 1

For $\frac{1}{a^2+x^2}$1/(a² + x²), with a > 0, substitute x = au, so dx = a du. The denominator becomes $a^2+a^2{u}^2=a^2\left(1+u^2\right)$a² + a²u² = a²(1 + u²), and the integral becomes $\int \frac{a}{a^2\left(1+u^2\right)}du=\left(\frac{1}{a}\right)\int \frac{1}{1+u^2}du=\left(\frac{1}{a}\right){\tan }^{-1}u+C$∫ a/(a²(1 + u²)) du = (1/a) ∫ 1/(1 + u²) du = (1/a) tan⁻¹u + C. Putting $u=\frac{x}{a}$u = x/a back: $\int \frac{1}{a^2+x^2}dx=\left(\frac{1}{a}\right){\tan }^{-1}\left(\frac{x}{a}\right)+C$∫ 1/(a² + x²) dx = (1/a) tan⁻¹(x/a) + C.

For $\frac{1}{9+x^2}$1/(9 + x²), a = 3, since 9 = 3 × 3. So the integral is $\left(\frac{1}{3}\right){\tan }^{-1}\left(\frac{x}{3}\right)+C$(1/3) tan⁻¹(x/3) + C. Check: differentiating gives $\left(\frac{1}{3}\right)\times \left(\frac{1}{3}\right)\times \frac{1}{1+\frac{x^2}{9}}=\left(\frac{1}{9}\right)\times \frac{9}{9+x^2}=\frac{1}{9+x^2}$(1/3) × (1/3) × 1/(1 + x²/9) = (1/9) × 9/(9 + x²) = 1/(9 + x²). From 0 to 3 the integral is $\left(\frac{1}{3}\right)\left(\frac{\pi }{4}\right)=\frac{\pi }{12}\approx 0.262$(1/3)(π/4) = π/12 ≈ 0.262.

For the root, the a cancels: $\sqrt{a^2-a^2{u}^2}=a\sqrt{1-u^2}$√(a² − a²u²) = a√(1 − u²), and dx = a du divided by that leaves $\frac{du}{\sqrt{1-u^2}}$du/√(1 − u²). So $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+C$∫ 1/√(a² − x²) dx = sin⁻¹(x/a) + C, with no $\frac{1}{a}$1/a in front. From 0 to 1, $\int \frac{1}{\sqrt{4-x^2}}dx={\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$∫ 1/√(4 − x²) dx = sin⁻¹½ = π/6.

A number in front of $x^2$x² is handled the same way. $1+4x^2=1+{\left(2x\right)}^2$1 + 4x² = 1 + (2x)², so $\int \frac{1}{1+4x^2}dx=\frac{1}{2}{\tan }^{-1}\left(2x\right)+C$∫ 1/(1 + 4x²) dx = ½ tan⁻¹(2x) + C, the ½ undoing the 2 the chain rule brings out. From 0 to ½ that is $\frac{1}{2}{\tan }^{-1}1=\frac{\pi }{8}\approx 0.393$½ tan⁻¹1 = π/8 ≈ 0.393.

The sign decides

$x^2-1$x² − 1 factors as (x − 1)(x + 1), so $\frac{1}{x^2-1}$1/(x² − 1) splits into partial fractions and integrates to logs: ½ ln|x − 1| − ½ ln|x + 1| + C. $x^2+1$x² + 1 is never 0, so it has no factors, and $\frac{1}{x^2+1}$1/(x² + 1) integrates to ${\tan }^{-1}x+C$tan⁻¹x + C instead.

The numerator matters too. $\frac{x}{1+x^2}$x/(1 + x²) is half of $\frac{2x}{1+x^2}$2x/(1 + x²), whose numerator is the derivative of its denominator, so it integrates to a log: $\frac{1}{2}\ln \left(1+x^2\right)+C$½ ln(1 + x²) + C. Only a constant over $1+x^2$1 + x² gives ${\tan }^{-1}$tan⁻¹.

The usual mistakes

Expecting a log. $\ln \left(1+x^2\right)$ln(1 + x²) differentiates to $\frac{2x}{1+x^2}$2x/(1 + x²), which has an x on top that $\frac{1}{1+x^2}$1/(1 + x²) does not.

Mixing up the pair. ${\tan }^{-1}$tan⁻¹ goes with $1+x^2$1 + x² and no root; ${\sin }^{-1}$sin⁻¹ goes with the root of $1-x^2$1 − x².

Leaving out the $\frac{1}{a}$1/a. ${\tan }^{-1}\left(\frac{x}{3}\right)$tan⁻¹(x/3) differentiates to $\frac{3}{9+x^2}$3/(9 + x²), three times too much.

Using $a^2$a² for a. In $\frac{1}{4+x^2}$1/(4 + x²), a = 2, so the answer is $\frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+C$½ tan⁻¹(x/2) + C, not $\frac{1}{4}{\tan }^{-1}\left(\frac{x}{4}\right)+C$¼ tan⁻¹(x/4) + C.

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