Hyperbolic and Tangent-Half-Angle Substitutions, calculus lesson

A square root like $\sqrt{x^2 + 1}$ or a fraction like $\frac{1}{2 + \cos x}$ looks stubborn until you find the right change of variable. Trigonometric substitution already handles many roots. Hyperbolic functions often do the same job with less fuss, and one clever substitution turns every rational expression in $\sin x$ and $\cos x$ into an ordinary rational function. Each trick comes with an interval where it is allowed, and ignoring that interval gives wrong answers.

Key terms

Hyperbolic substitutions

Definition Let $a > 0$ . The hyperbolic functions satisfy $\cosh^2 t - \sinh^2 t = 1$ , with $\cosh t > 0$ for all real $t$ and $\sinh t \ge 0$ for $t \ge 0$ .

For an integrand containing $\sqrt{a^2 + x^2}$ on any interval, substitute $x = a \sinh t$ with $t \in \mathbb{R}$ . Then $dx = a \cosh t \, dt$ and

\sqrt{a^2 + x^2} = \sqrt{a^2 (1 + \sinh^2 t)} = a \cosh t.

For an integrand containing $\sqrt{x^2 - a^2}$ on an interval where $x \ge a$ , substitute $x = a \cosh t$ with $t \ge 0$ . Then $dx = a \sinh t \, dt$ and $\sqrt{x^2 - a^2} = a \sinh t$ . On an interval where $x \le -a$ , first put $x = -u$ .

In plain words The identity $\cosh^2 t - \sinh^2 t = 1$ is built to cancel the expressions under these square roots, so the root disappears and you integrate something simpler.

Example To find the integral

\int \frac{dx}{(1 + x^2)^{3/2}},

put $x = \sinh t$ . Then $(1 + x^2)^{3/2} = \cosh^3 t$ and $dx = \cosh t \, dt$ , so

\int \frac{\cosh t \, dt}{\cosh^3 t} = \int \operatorname{sech}^2 t \, dt = \tanh t + C = \frac{x}{\sqrt{1 + x^2}} + C,

using $\tanh t = \sinh t / \cosh t$ and $\sinh t = x$ . Differentiating the result gives back the integrand, which confirms it.

Inverse hyperbolic forms

Definition The inverse hyperbolic functions have the logarithmic forms

\operatorname{arsinh} x = \ln\left(x + \sqrt{x^2 + 1}\right) \quad (x \in \mathbb{R}),

\operatorname{arcosh} x = \ln\left(x + \sqrt{x^2 - 1}\right) \quad (x \ge 1), \qquad \operatorname{artanh} x = \frac{1}{2} \ln \frac{1 + x}{1 - x} \quad (\lvert x \rvert < 1).

For $a > 0$ they give the antiderivatives

\int \frac{dx}{\sqrt{x^2 + a^2}} = \operatorname{arsinh} \frac{x}{a} + C \ \text{ on } \mathbb{R}, \qquad \int \frac{dx}{\sqrt{x^2 - a^2}} = \operatorname{arcosh} \frac{x}{a} + C \ \text{ on } (a, \infty),

\int \frac{dx}{a^2 - x^2} = \frac{1}{a} \operatorname{artanh} \frac{x}{a} + C \ \text{ on } (-a, a).

In plain words These are the answers that hyperbolic substitution produces, written either with inverse hyperbolic names or as plain logarithms. Both versions are correct, and they differ only by a constant when you expand the logarithm.

Example By the first formula with $a = 1$ ,

\int_0^{3/4} \frac{dx}{\sqrt{x^2 + 1}} = \operatorname{arsinh} \frac{3}{4} = \ln\left(\frac{3}{4} + \sqrt{\frac{9}{16} + 1}\right) = \ln\left(\frac{3}{4} + \frac{5}{4}\right) = \ln 2.

Tangent half-angle substitution

Definition On the interval $-\pi < x < \pi$ , the tangent half-angle (Weierstrass) substitution is $t = \tan \frac{x}{2}$ , so $x = 2 \arctan t$ with $t \in \mathbb{R}$ . It gives

\sin x = \frac{2t}{1 + t^2}, \qquad \cos x = \frac{1 - t^2}{1 + t^2}, \qquad dx = \frac{2 \, dt}{1 + t^2}.

The formulas for sine and cosine come from the double angle formulas

\sin x = \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}}, \qquad \cos x = \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}},

where the denominators equal $1$ . Dividing top and bottom by $\cos^2 \frac{x}{2}$ , which is not zero for $-\pi < x < \pi$ , gives the expressions in $t$ . The formula for $dx$ comes from differentiating $x = 2 \arctan t$ .

In plain words Sine, cosine and $dx$ all become fractions in $t$ with no trigonometry left, so the integral turns into an ordinary rational integral.

Example Over $0 \le x \le \pi/2$ the new limits are $t = \tan 0 = 0$ and $t = \tan \frac{\pi}{4} = 1$ . The denominator becomes

2 + \cos x = \frac{3 + t^2}{1 + t^2},

\int_0^{\pi/2} \frac{dx}{2 + \cos x} = \int_0^1 \frac{2 \, dt}{3 + t^2} = \frac{2}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}} \approx 0.6046.

Rationalizing trigonometric expressions

Definition A rational function of $\sin x$ and $\cos x$ is an expression $R(\sin x, \cos x)$ where $R(u, v)$ is a quotient of two polynomials in $u$ and $v$ . On any subinterval of $(-\pi, \pi)$ where $R(\sin x, \cos x)$ is defined, that is, where its denominator is not zero, the substitution $t = \tan \frac{x}{2}$ gives

\int R(\sin x, \cos x) \, dx = \int R\left(\frac{2t}{1 + t^2}, \frac{1 - t^2}{1 + t^2}\right) \frac{2}{1 + t^2} \, dt,

and the new integrand is a rational function of $t$ . So every such integral can be done with partial fractions.

In plain words No matter how messy the mix of sines and cosines, as long as it is built with only arithmetic, this substitution always leads to an integral you know how to finish.

Example In the integral of $1/(1 + \sin x + \cos x)$ over $[0, \pi/2]$ the denominator becomes

1 + \frac{2t}{1 + t^2} + \frac{1 - t^2}{1 + t^2} = \frac{2 + 2t}{1 + t^2}.

So the integrand times $dx$ is

\frac{1 + t^2}{2(1 + t)} \cdot \frac{2 \, dt}{1 + t^2} = \frac{dt}{1 + t},

and

\int_0^{\pi/2} \frac{dx}{1 + \sin x + \cos x} = \int_0^1 \frac{dt}{1 + t} = \ln 2.

The substitution is a fallback, not a first choice. To integrate $\sin^2 x \cos x$ the plain substitution $u = \sin x$ is much quicker.

Substitution intervals

Definition Substitution in a definite integral rests on this rule. If $g$ has a continuous derivative on $[\alpha, \beta]$ and $f$ is continuous on the range $g([\alpha, \beta])$ , then

\int_\alpha^\beta f(g(t)) \, g'(t) \, dt = \int_{g(\alpha)}^{g(\beta)} f(x) \, dx.

When you substitute $x = g(t)$ and want to return to $x$ at the end, $g$ must also be one to one on the interval, so that $t$ can be written as a function of $x$ . Every simplification of a root such as $\sqrt{\sinh^2 t + 1} = \cosh t$ or $\sqrt{\cosh^2 t - 1} = \sinh t$ must hold for the $t$ values used. For $t = \tan \frac{x}{2}$ this means working inside one interval $(-\pi + 2k\pi, \pi + 2k\pi)$ with $k$ an integer.

In plain words A substitution is only trustworthy on an interval where it behaves well. If the integration interval crosses a point where the substitution breaks, split the integral there.

Example The integrand $\frac{1}{2 + \cos x}$ is continuous and positive on $[0, 2\pi]$ , and

\int_0^{2\pi} \frac{dx}{2 + \cos x} = \frac{2\pi}{\sqrt{3}} \approx 3.6276.

The substitution suggests the antiderivative

F(x) = \frac{2}{\sqrt{3}} \arctan\left(\frac{\tan(x/2)}{\sqrt{3}}\right).

Its derivative is $1/(2 + \cos x)$ wherever it is defined, but it is undefined at $x = \pi$ and jumps there. Naively, $F(2\pi) - F(0) = 0 - 0 = 0$ , which is impossible for a positive integrand. Splitting at $\pi$ fixes it. Since the integrand is continuous, $\int_0^\pi$ is the limit of $F(b) - F(0)$ as $b \to \pi^-$ , and $\int_\pi^{2\pi}$ is the limit of $F(2\pi) - F(b)$ as $b \to \pi^+$ . As $x$ rises from $0$ toward $\pi$ , the values of $F$ rise from $0$ toward $\frac{2}{\sqrt{3}} \cdot \frac{\pi}{2}$ . Just after $\pi$ they start near $-\frac{2}{\sqrt{3}} \cdot \frac{\pi}{2}$ and rise to $0$ at $2\pi$ . Each half contributes $\frac{\pi}{\sqrt{3}}$ , for a total of $\frac{2\pi}{\sqrt{3}}$ .

Check yourself

Find $\int \frac{dx}{\sqrt{x^2 - 9}}$ for $x > 3$ .

Show the answer

With $x = 3\cosh t$ and $t > 0$ you get $\int dt = t + C$ , so the answer is $\operatorname{arcosh} \frac{x}{3} + C$ , or equivalently $\ln\left(x + \sqrt{x^2 - 9}\right) + C$ . The two forms differ by the constant $\ln 3$ .

Use $t = \tan \frac{x}{2}$ to evaluate $\int_0^{\pi/2} \frac{dx}{1 + \cos x}$ .

Show the answer

The substitution gives $1 + \cos x = \frac{2}{1 + t^2}$ , so the integrand times $dx$ is $\frac{1 + t^2}{2} \cdot \frac{2 \, dt}{1 + t^2} = dt$ . The limits are $0$ and $1$ , so the integral is $1$ .

Why can you not apply $t = \tan \frac{x}{2}$ directly to $\int_0^{3\pi/2} \frac{dx}{2 + \sin x}$ in one piece?

Show the answer

The interval contains $x = \pi$ , where $\tan \frac{x}{2}$ is undefined. Split the integral at $\pi$ . On each piece use the antiderivative from the substitution and take its one-sided limit at $\pi$ , just as in the $\frac{1}{2 + \cos x}$ example above.

Hyperbolic and Tangent-Half-Angle Substitutions

Theory, part 117

Practice, part 118

Key terms

Hyperbolic substitutions

Inverse hyperbolic forms

Tangent half-angle substitution

Rationalizing trigonometric expressions

Substitution intervals

Check yourself

How this lesson connects

Builds on