Trigonometric Substitution for Integrals

How trigonometric substitution works

1.Trigonometric substitution is a method for evaluating integrals that contain square roots of quadratic expressions. Certain algebraic forms become considerably easier to handle once they are rewritten through the Pythagorean identities of trigonometry. The method consists in a change of variables $x = \phi(\theta)$ chosen so that the quadratic expression under the radical is transformed into the square of a trigonometric function in the new variable.

2.After an appropriate algebraic manipulation, many integrals in elementary calculus can be reduced to one of the following canonical forms, with $a > 0$:

$$\sqrt{a^2 - x^2} \qquad \sqrt{x^2 + a^2} \qquad \sqrt{x^2 - a^2}$$

3.For the radical to be real-valued, the variable $x$ must lie in the corresponding domain:

5.These restrictions are essential, since they determine not only where the integrand is defined, but also the admissible range of the auxiliary angle $\theta$ introduced in the substitution. In particular, the choice of the interval for $\theta$ ensures that inverse trigonometric functions are well defined and that absolute values arising from square roots can be handled consistently. Each of these expressions is naturally linked to a Pythagorean identity:

$$1 - \sin^2\theta = \cos^2\theta$$

$$1 + \tan^2\theta = \sec^2\theta$$

$$\sec^2\theta - 1 = \tan^2\theta$$

6.

The substitution in each case is chosen so that the term inside the square root matches the left-hand side of one of these identities, which turns the radical into an expression without radicals. Integrals involving rational functions of $\sin x$ and $\cos x$ are instead usually treated through the Weierstrass substitution, which transforms trigonometric expressions into rational functions of a new variable.

7.In practice, the expression under the square root is rarely presented in one of the three canonical forms straight away. A common preliminary step is to rewrite a general quadratic $ax^2 + bx + c$ in a form that matches one of the standard patterns, by completing the square. For instance, $x^2 + 4x + 5$ becomes $(x + 2)^2 + 1$ after completing the square, which is of the form $u^2 + a^2$ with $u = x + 2$ and $a = 1$. Once the quadratic has been rewritten in this way, a simple substitution $u = x + k$ reduces the integral to one of the three cases described below, and the appropriate trigonometric substitution can then be applied.

8.

Recognising this preliminary step often determines which substitution is needed. When the integrand does not immediately match a familiar pattern, the rewriting by completing the square is what makes the correct case visible.

From substitutions to geometry

9.From a geometric point of view, these substitutions can be interpreted as parametrisations of conic sections. The correspondence between each Pythagorean identity and the curve it describes is the following.

11.Trigonometric substitution can therefore be understood as a geometric reparametrisation of quadratic curves, and not only as an algebraic device.

The form $\sqrt{a^2 - x^2}$

12.When the integrand contains an expression of the form $\sqrt{a^2 - x^2}$, the appropriate trigonometric substitution reflects the Pythagorean identity $1 - \sin^2\theta = \cos^2\theta$, which itself follows from the fundamental relation between sine and cosine. Set:

$$x = a\sin\theta$$

13.so that the algebraic quantity under the square root can be rewritten in terms of a trigonometric function. Differentiating both sides with respect to $\theta$:

$$dx = a\cos\theta \ d\theta$$

14.Substituting $x = a\sin\theta$ into the radical expression:

$$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta}$$

15.Factoring $a^2$ from the expression inside the radical yields:

$$\sqrt{a^2(1 - \sin^2\theta)}$$

16.Using the identity $1 - \sin^2\theta = \cos^2\theta$, this becomes:

$$\sqrt{a^2\cos^2\theta} = a\sqrt{\cos^2\theta} = a|\cos\theta|$$

17.To avoid ambiguity related to the absolute value, the angle $\theta$ is customarily restricted to the interval:

$$\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

18.since on this interval $\cos\theta \geq 0$. Under this restriction, the absolute value is no longer necessary, and the radical simplifies to:

$$\sqrt{a^2 - x^2} = a\cos\theta$$

19.Returning to the original variable $x$, the relationships implied by the substitution can be written explicitly as:

$$\sin\theta = \frac{x}{a} \qquad \cos\theta = \frac{\sqrt{a^2 - x^2}}{a}$$

20.The angle can be expressed through the arcsine:

$$\theta = \arcsin\left(\frac{x}{a}\right)$$

21.These relations allow the final result of the integration to be expressed entirely in terms of the original variable.

22.

A geometric interpretation is often helpful. If $\sin\theta = x/a$, a right triangle with hypotenuse $a$, opposite side $x$, and adjacent side $\sqrt{a^2 - x^2}$ encodes the relevant identities.

23.The standard correspondences for $\sqrt{a^2 - x^2}$ can be summarised as follows:

Geometric interpretation

25.When working with trigonometric substitutions, it is often useful to visualise the relationship between $\theta$ and $x$ through a right triangle. Reading the values of the trigonometric functions directly from the sides of the triangle removes the need to solve for $\theta$ explicitly.

26.From $x = a\sin\theta$, a right triangle is constructed where:

28.Since $\sin\theta = x/a$, it follows that $\cos\theta = \sqrt{a^2 - x^2}/a$. This geometric representation allows all trigonometric functions of $\theta$ to be rewritten directly in terms of $x$.

29.

The same construction applies to the other two standard forms. For $\sqrt{x^2 + a^2}$ the triangle has opposite side $x$, adjacent side $a$, and hypotenuse $\sqrt{x^2 + a^2}$; for $\sqrt{x^2 - a^2}$ the hypotenuse becomes $x$, the adjacent side $a$, and the opposite side $\sqrt{x^2 - a^2}$. In each case the triangle is built directly from the substitution and serves as a guide for the back-substitution step.

Example 1

30.Evaluate the following integral:

$$\int \sqrt{a^2 - x^2} \ dx \qquad (a > 0)$$

31.Since the integrand contains the expression $\sqrt{a^2 - x^2}$, introduce the trigonometric substitution:

$$x = a\sin\theta \qquad \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

32.so that the identity $1 - \sin^2\theta = \cos^2\theta$ applies with $\cos\theta \geq 0$ on this interval. Differentiating the substitution gives:

$$dx = a\cos\theta \ d\theta$$

33.The radical rewrites as:

$$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\sqrt{1 - \sin^2\theta} = a\cos\theta$$

34.Substituting everything into the integral:

$$ \begin{align} \int \sqrt{a^2 - x^2} \ dx &= \int (a\cos\theta)(a\cos\theta \ d\theta) \\[6pt] &= a^2 \int \cos^2\theta \ d\theta \end{align} $$

35.To integrate $\cos^2\theta$, use the double-angle identity:

$$\cos^2\theta = \frac{1 + \cos 2\theta}{2}$$

36.Therefore:

$$a^2 \int \cos^2\theta \ d\theta = \frac{a^2}{2}\int (1 + \cos 2\theta) \ d\theta$$

37.Integrating term by term:

$$\frac{a^2}{2}\theta + \frac{a^2}{4}\sin 2\theta + c$$

38.Returning to the variable $x$, the substitution $x = a\sin\theta$ gives:

$$\theta = \arcsin\left(\frac{x}{a}\right)$$

39.Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$ together with:

$$\sin\theta = \frac{x}{a} \qquad \cos\theta = \frac{\sqrt{a^2 - x^2}}{a}$$

40.one obtains:

$$\sin 2\theta = 2 \cdot \frac{x}{a} \cdot \frac{\sqrt{a^2 - x^2}}{a} = \frac{2x\sqrt{a^2 - x^2}}{a^2}$$

41.Substituting back:

$$\frac{a^2}{2}\theta + \frac{a^2}{4}\sin 2\theta = \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) + \frac{x}{2}\sqrt{a^2 - x^2}$$

42.The result is:

$$\int \sqrt{a^2 - x^2} \ dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) + c$$

43.

The integral has been reduced to a trigonometric form, evaluated through standard identities, and finally rewritten entirely in terms of the original variable $x$.

The form $\sqrt{x^2 + a^2}$

44.When the integrand contains an expression of the form $\sqrt{x^2 + a^2}$, a substitution based on the Pythagorean identity:

$$1 + \tan^2\theta = \sec^2\theta$$

45.is convenient. This identity relates the tangent and the secant. Set $x = a\tan\theta$ so that the quadratic expression inside the square root can be rewritten in terms of a trigonometric function. Differentiating both sides with respect to $\theta$:

$$dx = a\sec^2\theta \ d\theta$$

46.Substituting $x = a\tan\theta$ into the radical expression:

$$\sqrt{x^2 + a^2} = \sqrt{a^2\tan^2\theta + a^2}$$

47.Factoring $a^2$ from the expression inside the square root:

$$\sqrt{a^2(\tan^2\theta + 1)}$$

48.Using the identity $1 + \tan^2\theta = \sec^2\theta$, this becomes:

$$\sqrt{a^2\sec^2\theta} = a\sqrt{\sec^2\theta} = a|\sec\theta|$$

49.To eliminate the ambiguity introduced by the absolute value, the angle $\theta$ is restricted to:

$$\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

50.since on this interval $\cos\theta > 0$ and therefore $\sec\theta > 0$. Under this restriction, the radical simplifies to:

$$\sqrt{x^2 + a^2} = a\sec\theta$$

51.Returning to the original variable $x$, the relationships implied by the substitution can be written explicitly as:

$$\tan\theta = \frac{x}{a} \qquad \sec\theta = \frac{\sqrt{x^2 + a^2}}{a} \qquad \theta = \arctan\left(\frac{x}{a}\right)$$

52.

The geometric interpretation is immediate. From the relation $\tan\theta = x/a$, a right triangle can be constructed in which the adjacent side has length $a$, the opposite side has length $x$, and the hypotenuse, by the Pythagorean theorem, has length $\sqrt{x^2 + a^2}$. The triangle provides a geometric picture of the substitution and clarifies why the radical expression collapses into a trigonometric function.

53.The standard correspondences for $\sqrt{x^2 + a^2}$ can be summarised as follows:

Example 2

55.Evaluate the following integral:

$$\int \frac{dx}{\sqrt{x^2 + a^2}} \qquad (a > 0)$$

56.Since the integrand contains $\sqrt{x^2 + a^2}$, introduce the trigonometric substitution:

$$x = a\tan\theta \qquad \theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$

57.so that the identity $1 + \tan^2\theta = \sec^2\theta$ applies with $\sec\theta > 0$ on this interval. Differentiating the substitution gives:

$$dx = a\sec^2\theta \ d\theta$$

58.The radical rewrites as:

$$ \begin{align} \sqrt{x^2 + a^2} &= \sqrt{a^2\tan^2\theta + a^2} \\[6pt] &= a\sqrt{\tan^2\theta + 1} \\[6pt] &= a\sec\theta \end{align} $$

59.Substituting everything into the integral:

$$\int \frac{dx}{\sqrt{x^2 + a^2}} = \int \frac{a\sec^2\theta}{a\sec\theta} \ d\theta = \int \sec\theta \ d\theta$$

60.To evaluate $\int \sec\theta \ d\theta$, multiply the integrand by $(\sec\theta + \tan\theta)/(\sec\theta + \tan\theta)$:

$$ \begin{align} \int \sec\theta \ d\theta &= \int \frac{\sec\theta(\sec\theta + \tan\theta)}{\sec\theta + \tan\theta} \ d\theta \\[6pt] &= \int \frac{\sec^2\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta} \ d\theta \end{align} $$

61.Setting $u = \sec\theta + \tan\theta$ gives $du = (\sec^2\theta + \sec\theta\tan\theta) \ d\theta$, so the numerator is exactly $du$ and the integral reduces to:

$$\int \frac{du}{u} = \ln|u| + c = \ln|\sec\theta + \tan\theta| + c$$

62.Returning to the variable $x$, using:

$$\tan\theta = \frac{x}{a} \qquad \sec\theta = \frac{\sqrt{x^2 + a^2}}{a}$$

63.one obtains:

$$\ln|\sec\theta + \tan\theta| + c = \ln\left|\frac{\sqrt{x^2 + a^2} + x}{a}\right| + c$$

64.Since $\sqrt{x^2 + a^2} > |x|$ for every $x \in \mathbb{R}$, the quantity $\sqrt{x^2 + a^2} + x$ is strictly positive, and the absolute value can be dropped. Moreover:

$$\ln\left|\frac{\sqrt{x^2 + a^2} + x}{a}\right| = \ln\!\left(\sqrt{x^2 + a^2} + x\right) - \ln a$$

65.and $\ln a$ is a constant that can be absorbed into $c$. The result is:

$$\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\!\left(\sqrt{x^2 + a^2} + x\right) + c$$

66.

The integral has been reduced to a trigonometric form, evaluated through a standard manipulation of the secant, and finally rewritten entirely in terms of the original variable $x$.

The form $\sqrt{x^2 - a^2}$

67.When the integrand contains an expression of the form $\sqrt{x^2 - a^2}$, the natural substitution is based on the Pythagorean identity:

$$\sec^2\theta - 1 = \tan^2\theta$$

68.which is equivalent to the fundamental relation $1 + \tan^2\theta = \sec^2\theta$. Set:

$$x = a\sec\theta$$

69.so that the quadratic expression inside the square root can be rewritten in terms of trigonometric functions. Differentiating both sides with respect to $\theta$:

$$dx = a\sec\theta\tan\theta \ d\theta$$

70.Substituting $x = a\sec\theta$ into the radical expression:

$$\sqrt{x^2 - a^2} = \sqrt{a^2\sec^2\theta - a^2}$$

71.Factoring $a^2$ inside the square root:

$$\sqrt{a^2(\sec^2\theta - 1)}$$

72.Using the identity $\sec^2\theta - 1 = \tan^2\theta$, this becomes:

$$\sqrt{a^2\tan^2\theta} = a\sqrt{\tan^2\theta} = a|\tan\theta|$$

73.The presence of the absolute value reflects the fact that the sign of $\tan\theta$ depends on the chosen domain for $\theta$. Working under the assumption $x \geq a$, a convenient restriction is:

$$\theta \in \left[0, \frac{\pi}{2}\right)$$

74.since on this interval $\sec\theta \geq 1$ and $\tan\theta \geq 0$. Under this restriction, the radical simplifies to:

$$\sqrt{x^2 - a^2} = a\tan\theta$$

75.

When $x \leq -a$, the corresponding restriction is $\theta \in \left(\frac{\pi}{2}, \pi\right]$, so that $\sec\theta \leq -1$ and $\tan\theta \leq 0$; in that case $|\tan\theta| = -\tan\theta$. For most textbook problems it suffices to assume $x \geq a$ and work with the interval $\left[0, \frac{\pi}{2}\right)$.

76.Returning to the original variable $x$, the relationships implied by the substitution can be written explicitly as:

$$\sec\theta = \frac{x}{a} \qquad \tan\theta = \frac{\sqrt{x^2 - a^2}}{a}$$

77.Equivalently, the angle itself can be expressed through an inverse trigonometric function:

$$\theta = \operatorname{arcsec}\left(\frac{x}{a}\right)$$

78.on a suitable domain. In many practical situations, however, it is sufficient to rewrite $\tan\theta$ and $\sec\theta$ directly in terms of $x$ and $\sqrt{x^2 - a^2}$, without explicitly solving for $\theta$.

79.

The geometric interpretation follows directly from the relation $\sec\theta = x/a$. A right triangle can be constructed in which the hypotenuse has length $x$, the adjacent side has length $a$, and the opposite side, by the Pythagorean theorem, has length $\sqrt{x^2 - a^2}$. This triangle makes the substitution geometrically transparent and provides a way to read off the values of $\sec\theta$ and $\tan\theta$ directly from the sides, without solving for $\theta$ explicitly.

Example 3

80.Evaluate the following integral:

$$\int \frac{dx}{\sqrt{x^2 - a^2}} \qquad (a > 0, \ x > a)$$

81.Since the integrand contains $\sqrt{x^2 - a^2}$, introduce the trigonometric substitution:

$$x = a\sec\theta \qquad \theta \in \left[0, \frac{\pi}{2}\right)$$

82.so that the identity $\sec^2\theta - 1 = \tan^2\theta$ applies with $\tan\theta \geq 0$ on this interval. Differentiating the substitution gives:

$$dx = a\sec\theta\tan\theta \ d\theta$$

83.The radical rewrites as:

$$ \begin{align} \sqrt{x^2 - a^2} &= \sqrt{a^2\sec^2\theta - a^2} \\[6pt] &= a\sqrt{\sec^2\theta - 1} \\[6pt] &= a\tan\theta \end{align} $$

84.Substituting everything into the integral:

$$\int \frac{dx}{\sqrt{x^2 - a^2}} = \int \frac{a\sec\theta\tan\theta}{a\tan\theta} \ d\theta = \int \sec\theta \ d\theta$$

85.Evaluating $\int \sec\theta \ d\theta$ as in Example 2:

$$\int \sec\theta \ d\theta = \ln|\sec\theta + \tan\theta| + c$$

86.Returning to the variable $x$, using:

$$\sec\theta = \frac{x}{a} \qquad \tan\theta = \frac{\sqrt{x^2 - a^2}}{a}$$

87.one obtains:

$$\ln|\sec\theta + \tan\theta| + c = \ln\!\left|\frac{x + \sqrt{x^2 - a^2}}{a}\right| + c$$

88.Since $x > a > 0$ and $\sqrt{x^2 - a^2} \geq 0$, the quantity $x + \sqrt{x^2 - a^2}$ is strictly positive, and the absolute value can be dropped. Moreover:

$$\ln\!\left(\frac{x + \sqrt{x^2 - a^2}}{a}\right) = \ln\!\left(x + \sqrt{x^2 - a^2}\right) - \ln a$$

89.and $\ln a$ is a constant that can be absorbed into $c$. The result is:

$$\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\!\left(x + \sqrt{x^2 - a^2}\right) + c$$

90.

The structure of the derivation closely mirrors that of Example 2. In both cases the substitution reduces the integral to $\int \sec\theta \ d\theta$, and the difference lies entirely in the back-substitution step, where the expressions for $\sec\theta$ and $\tan\theta$ in terms of $x$ reflect the geometry of the two distinct radical forms.

Example 4

91.The previous examples all began with a radical already written in one of the three canonical forms. In practice the quadratic under the square root is often a general trinomial, and the substitution becomes visible only after completing the square. Consider the following integral:

$$\int \frac{dx}{\sqrt{x^2 + 4x + 5}}$$

92.The expression under the radical is not yet in a canonical form. The first step is to complete the square in the quadratic:

$$x^2 + 4x + 5 = (x + 2)^2 + 1$$

93.This rewriting shows that the radical has the form $\sqrt{u^2 + a^2}$ with $u = x + 2$ and $a = 1$. Introduce the auxiliary substitution:

$$u = x + 2 \qquad du = dx$$

94.which transforms the integral into:

$$\int \frac{dx}{\sqrt{x^2 + 4x + 5}} = \int \frac{du}{\sqrt{u^2 + 1}}$$

95.The integral on the right is exactly the canonical form treated in Example 2, with $a = 1$. Applying the trigonometric substitution $u = \tan\theta$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the same derivation gives:

$$\int \frac{du}{\sqrt{u^2 + 1}} = \ln\!\left(\sqrt{u^2 + 1} + u\right) + c$$

96.Substituting $u = x + 2$ back into the result:

$$\int \frac{dx}{\sqrt{x^2 + 4x + 5}} = \ln\!\left(\sqrt{x^2 + 4x + 5} + x + 2\right) + c$$

97.

The decisive step is the initial completion of the square, which exposes the canonical form hidden inside the general quadratic. Once the radical has been rewritten as $\sqrt{u^2 + a^2}$, the problem reduces to a case already solved, and the trigonometric substitution proceeds as before.

Decision procedure

98.The following stepwise procedure summarises the application of trigonometric substitution to a generic integral containing a radical of a quadratic.

100.

When the integrand is a rational function of $x$ rather than the radical of a quadratic, partial fraction decomposition and the techniques on the page on integrals of rational functions are the appropriate tools. For rational functions of $\sin x$ and $\cos x$, the Weierstrass substitution provides a systematic alternative.