The Weierstrass Substitution

When to use the substitution

1.The Weierstrass substitution is a method for evaluating integrals that cannot be computed directly, whose integrands are rational functions involving sine and cosine, by making a suitable change of variable. In general, integrals of this type have forms similar to those listed below, where the numerator and denominator are polynomials in $\sin x$ and $\cos x:$

$$\frac{1}{1 + \sin x} \qquad \frac{1}{5 - 3\cos x} \qquad \frac{1}{\sin x + \cos x} \tag{1}$$

2.We have already covered some techniques for integrating trigonometric functions, based on power-reduction formulas or separating out factors. However, in cases such as those in $(1)$ these techniques do not provide a useful way to obtain a result readily. In the cases we will consider shortly, we need to make a change of variable that reduces the problem to integrating a simple rational function, which is generally handled using polynomial division and partial fraction decomposition.

3.The Weierstrass substitution is based on introducing the following variable:

$$t = \tan\left(\frac{x}{2}\right) \tag{2}$$

4.This substitution allows us to write $\sin x,$ $\cos x$ and the differential $dx$ as rational expressions in $t.$ To derive these expressions, as we will see shortly, we use the double-angle formulas and the Pythagorean identity. Once we have calculated the antiderivative in $t,$ we return to the original variable $x$ by substituting $t = \tan(x/2).$

5.First, we examine how to derive the expressions for sine, cosine and the differential. I should mention at the outset that recalling the various trigonometric identities we will use in the calculations will require a little effort, so I encourage you to consult the relevant page before proceeding. For sine, we start with the double-angle formula, which gives:

$$\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) \tag{3}$$

6.Multiplying by $\cos(x/2) / \cos(x/2)$ allows us to rewrite the right-hand side as:

$$2\tan\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right) \tag{4}$$

7.The following trigonometric identity holds for cosine:

$$\cos^2\left(\frac{x}{2}\right) = \frac{1}{1 + \tan^2(x/2)}$$

8.Substituting this into $(4)$ allows us to rewrite the expression for sine in $(3)$ as:

$$\sin x = \frac{2t}{1 + t^2} \tag{5}$$

9.A similar procedure applies to deriving the expression for cosine, using the corresponding double-angle formula:

$$\cos x = \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right)$$

10.We divide the right-hand side by $\cos^2(x/2) + \sin^2(x/2)$ and then divide the numerator and denominator by $\cos^2(x/2),$ obtaining:

$$\cos x = \frac{1 - t^2}{1 + t^2} \tag{6}$$

11.Finally, we calculate the differential by differentiating relation $(2)$ with respect to $x.$ This gives:

$$\frac{dt}{dx} = \frac{1}{2}\sec^2\left(\frac{x}{2}\right) = \frac{1}{2}\left(1 + \tan^2\left(\frac{x}{2}\right)\right) = \frac{1 + t^2}{2}$$

12.Solving for $dx$ gives the third identity on which the substitution is based, namely:

$$dx = \frac{2}{1 + t^2} \ dt \tag{7}$$

13.The formulas $(5),$ $(6)$ and $(7)$ just derived therefore allow us to apply the Weierstrass method by substituting them into the integral of any rational expression in $\sin x$ and $\cos x$ to express it in the simpler variable $t$. To recap, the substitutions to apply are as follows:

Practical applications

15.We work through a few examples to illustrate how the Weierstrass method is applied in practice. Consider the following integral, whose integrand is a rational function of sine:

$$\int \frac{dx}{1 + \sin x}$$

16.Using identity $(5),$ we can rewrite the denominator as follows:

$$ \begin{align} 1 + \sin x &= 1 + \frac{2t}{1+t^2} \\[6pt] &= \frac{1 + t^2 + 2t}{1+t^2} \\[6pt] &= \frac{(1+t)^2}{1+t^2} \end{align} $$

17.We now rewrite the differential using $(7),$ obtaining:

$$ \begin{align} \frac{1}{1+\sin x} \ dx &= \frac{1+t^2}{(1+t)^2} \cdot \frac{2}{1+t^2} \ dt \\[6pt] &= \frac{2}{(1+t)^2} \ dt \end{align} $$

18.This gives an integrand whose antiderivative can be calculated directly:

$$\int \frac{2}{(1+t)^2} \ dt = -\frac{2}{1+t} + c$$

19.At this point, returning to the original variable through the identity $t = \tan(x/2),$ we obtain the result:

$$\int \frac{dx}{1 + \sin x} = -\frac{2}{1 + \tan(x/2)} + c$$

20.Now consider the following integral, whose integrand is a rational function of cosine:

$$\int \frac{dx}{5 - 3\cos x}$$

21.Using identity $(6),$ we can rewrite the denominator as:

$$ \begin{align} 5 - 3\cos x &= 5 - 3 \cdot \frac{1 - t^2}{1+t^2} \\[6pt] &= \frac{5(1+t^2) - 3(1-t^2)}{1+t^2} \\[6pt] &= \frac{2 + 8t^2}{1+t^2} \\[8pt] &= \frac{2(1 + 4t^2)}{1+t^2} \end{align} $$

22.Now substituting the expression for the differential using $(7),$ we obtain:

$$ \begin{align} \frac{1}{5-3\cos x} \ dx &= \frac{1+t^2}{2(1+4t^2)} \cdot \frac{2}{1+t^2} \ dt \\[6pt] &= \frac{dt}{1 + 4t^2} \end{align} $$

23.Here too, the remaining integral reduces to an elementary form, and since $1 + 4t^2 = 1 + (2t)^2,$ we set $u = 2t$ and $du = 2 \ dt,$ so the integral becomes:

$$ \begin{align} \int \frac{dt}{1 + 4t^2} &= \frac{1}{2}\int \frac{du}{1 + u^2} \\[6pt] &= \frac{1}{2}\arctan u + c \\[6pt] &= \frac{1}{2}\arctan(2t) + c \end{align} $$

24.Returning to the variable $x,$ we obtain the antiderivative:

$$\int \frac{dx}{5 - 3\cos x} = \frac{1}{2}\arctan(2\tan(x/2)) + c$$

25.We present one final case by evaluating the following integral:

$$\int \frac{dx}{2 + \sin x}$$

26.Applying the identities already derived, we can rewrite the denominator as follows:

$$ \begin{align} 2 + \sin x &= 2 + \frac{2t}{1+t^2} \\[6pt] &= \frac{2(1+t^2) + 2t}{1+t^2} \\[6pt] &= \frac{2(t^2 + t + 1)}{1+t^2} \end{align} $$

27.The expression to be integrated therefore becomes:

$$ \begin{align} \frac{1}{2+\sin x} \ dx &= \frac{1+t^2}{2(t^2+t+1)} \cdot \frac{2}{1+t^2} \ dt \\[6pt] &= \frac{dt}{t^2+t+1} \end{align} $$

28.Completing the square in the denominator, we obtain:

$$t^2 + t + 1 = \left(t + \frac{1}{2}\right)^2 + \frac{3}{4}$$

29.The integral therefore becomes:

$$\int \frac{dt}{\left(t+\frac{1}{2}\right)^2 + \frac{3}{4}}$$

30.With a little practice, we can recognize that this expression matches the elementary form:

$$\int \frac{du}{u^2 + a^2}$$

31.In this case, we have $u = t + 1/2$ and $a = \sqrt{3}/2.$ Using the corresponding integration formula, we obtain:

$$\int \frac{dt}{t^2+t+1} = \frac{2}{\sqrt{3}}\arctan\left(\frac{2t+1}{\sqrt{3}}\right) + c$$

32.Once again substituting $t = \tan(x/2),$ we obtain the final expression:

$$\int \frac{dx}{2+\sin x} = \frac{2}{\sqrt{3}}\arctan\left(\frac{2\tan(x/2)+1}{\sqrt{3}}\right) + c$$

33.

Here too, we have seen that after the substitution, the integrand becomes a rational function of $t,$ simpler than the original one, and its antiderivative may contain arctangents and logarithms, which typically appear when integrating rational functions.

Domain conditions

34.We need to consider the domains of the functions involved in the substitutions. The substitution $t = \tan(x/2)$ is defined for every $x$ such that $x/2 \neq \pi/2 + k\pi,$ that is, for every $x \notin \pi + 2\pi\mathbb{Z}.$ When evaluating an indefinite integral, the resulting formulas hold on intervals where both the original integrand and the substitution are defined.

35.Greater care is needed when applying the substitution to a definite integral whose limits of integration may lie in different intervals. In this case, we apply the substitution separately to each part of the domain of integration on which it is defined, has a continuous nonzero derivative and is invertible on the interval in question, then add the resulting contributions. Mechanically applying the substitutions above across a point of the form $x = (2k+1)\pi$ could lead to incorrect results, since the substitution might not be defined at that point. Consider, for example, the definite integral:

$$\int_0^{2\pi} \frac{dx}{5 - 3\cos x}$$

36.The integrand is continuous throughout the interval $[0,2\pi]$ but the substitution $t = \tan(x/2),$ by contrast, is not defined at $x = \pi.$ For this reason, we must split the integral at this point and rewrite the two contributions using the following limits:

$$ \lim_{a \to \pi^-}\int_0^a \frac{dx}{5 - 3\cos x} + \lim_{b \to \pi^+}\int_b^{2\pi} \frac{dx}{5 - 3\cos x} $$

37.As we have already shown in the preceding examples, the Weierstrass substitution transforms the expression to be integrated into:

$$\frac{dx}{5 - 3\cos x} = \frac{dt}{1 + 4t^2}$$

38.On the first interval, $x = 0$ corresponds to $t = 0,$ while $t \to +\infty$ as $x \to \pi^-.$ On the second interval, $t \to -\infty$ as $x \to \pi^+,$ while $x = 2\pi$ corresponds to $t = 0.$ We therefore obtain two improper integrals, which we write as follows:

$$\int_0^{+\infty} \frac{dt}{1 + 4t^2} + \int_{-\infty}^0 \frac{dt}{1 + 4t^2}$$

39.We calculate the two contributions separately and obtain the following values for the respective integrals:

$$ \begin{align} \int_0^{+\infty} \frac{dt}{1 + 4t^2} &= \lim_{A \to +\infty}\left[\frac{1}{2}\arctan(2t)\right]_0^A = \frac{\pi}{4} \\[6pt] \int_{-\infty}^0 \frac{dt}{1 + 4t^2} &= \lim_{B \to -\infty}\left[\frac{1}{2}\arctan(2t)\right]_B^0 = \frac{\pi}{4} \end{align} $$

40.Adding the two contributions, we finally obtain the value of the original integral:

$$\frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$$

41.

The given integral therefore equals $\pi/2.$ Transforming only the original limits of integration $0$ and $2\pi$ would have produced two limits both equal to zero, failing to account for the passage through the point where the substitution is undefined.

Final remarks

42.The Weierstrass substitution is not always the most efficient method. When the integrand contains only even powers of $\sin x$ and $\cos x$ it is preferable to use the basic methods for integrating trigonometric functions. The same also applies when the integrand can be rewritten using the following identities:

43.$$\sin^2 x = (1 - \cos 2x)/2$$ $$\cos^2 x = (1 + \cos 2x)/2$$

44.When the integrand has the form $R(\sin x)\cos x$ or $R(\cos x)\sin x,$ the direct substitutions $u = \sin x$ or $u = \cos x$ are quicker. As a practical guideline, it is advisable to use the Weierstrass substitution only when the integrand is a rational function of $\sin x$ and $\cos x$ with no obvious simplifications available through trigonometric identities or direct substitutions. In other cases, simpler techniques lead to shorter calculations.