Trigonometric substitution

Short description: Technique of integral evaluation

In mathematics, trigonometric substitution is the replacement of trigonometric functions for other expressions. In calculus, trigonometric substitution is a technique for evaluating integrals. Moreover, one may use the trigonometric identities to simplify certain integrals containing radical expressions. Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.

Case I: Integrands containing a² − x²

Let [math]\displaystyle{ x = a \sin \theta, }[/math] and use the identity [math]\displaystyle{ 1-\sin^2 \theta = \cos^2 \theta. }[/math]

Examples of Case I

Geometric construction for Case I

Example 1

In the integral

[math]\displaystyle{ \int\frac{dx}{\sqrt{a^2-x^2}}, }[/math]

we may use

[math]\displaystyle{ x=a\sin \theta,\quad dx=a\cos\theta\, d\theta, \quad \theta=\arcsin\frac{x}{a}. }[/math]

Then, [math]\displaystyle{ \begin{align} \int\frac{dx}{\sqrt{a^2-x^2}} &= \int\frac{a\cos\theta \,d\theta}{\sqrt{a^2-a^2\sin^2 \theta}} \\[6pt] &= \int\frac{a\cos\theta\,d\theta}{\sqrt{a^2(1 - \sin^2 \theta )}} \\[6pt] &= \int\frac{a\cos\theta\,d\theta}{\sqrt{a^2\cos^2\theta}} \\[6pt] &= \int d\theta \\[6pt] &= \theta + C \\[6pt] &= \arcsin\frac{x}{a}+C. \end{align} }[/math]

The above step requires that [math]\displaystyle{ a \gt 0 }[/math] and [math]\displaystyle{ \cos \theta \gt 0. }[/math] We can choose [math]\displaystyle{ a }[/math] to be the principal root of [math]\displaystyle{ a^2, }[/math] and impose the restriction [math]\displaystyle{ -\pi /2 \lt \theta \lt \pi /2 }[/math] by using the inverse sine function.

For a definite integral, one must figure out how the bounds of integration change. For example, as [math]\displaystyle{ x }[/math] goes from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ a/2, }[/math] then [math]\displaystyle{ \sin \theta }[/math] goes from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ 1/2, }[/math] so [math]\displaystyle{ \theta }[/math] goes from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ \pi / 6. }[/math] Then,

[math]\displaystyle{ \int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}=\int_0^{\pi/6} d\theta = \frac{\pi}{6}. }[/math]

Some care is needed when picking the bounds. Because integration above requires that [math]\displaystyle{ -\pi /2 \lt \theta \lt \pi /2 }[/math] , [math]\displaystyle{ \theta }[/math] can only go from [math]\displaystyle{ 0 }[/math] to [math]\displaystyle{ \pi / 6. }[/math] Neglecting this restriction, one might have picked [math]\displaystyle{ \theta }[/math] to go from [math]\displaystyle{ \pi }[/math] to [math]\displaystyle{ 5\pi /6, }[/math] which would have resulted in the negative of the actual value.

Alternatively, fully evaluate the indefinite integrals before applying the boundary conditions. In that case, the antiderivative gives

[math]\displaystyle{ \int_{0}^{a/2} \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \left( \frac{x}{a} \right) \Biggl|_{0}^{a/2} = \arcsin \left ( \frac{1}{2}\right) - \arcsin (0) = \frac{\pi}{6} }[/math] as before.

Example 2

The integral

[math]\displaystyle{ \int\sqrt{a^2-x^2}\,dx, }[/math]

may be evaluated by letting [math]\displaystyle{ x=a\sin \theta,\, dx=a\cos\theta\, d\theta,\, \theta=\arcsin\frac{x}{a}, }[/math] where [math]\displaystyle{ a \gt 0 }[/math] so that [math]\displaystyle{ \sqrt{a^2}=a, }[/math] and [math]\displaystyle{ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2} }[/math] by the range of arcsine, so that [math]\displaystyle{ \cos \theta \ge 0 }[/math] and [math]\displaystyle{ \sqrt{\cos^2 \theta} = \cos \theta. }[/math]

Then, [math]\displaystyle{ \begin{align} \int\sqrt{a^2-x^2}\,dx &= \int\sqrt{a^2-a^2\sin^2\theta}\,(a\cos\theta) \,d\theta \\[6pt] &= \int\sqrt{a^2(1-\sin^2\theta)}\,(a\cos\theta) \,d\theta \\[6pt] &= \int\sqrt{a^2(\cos^2\theta)}\,(a\cos\theta) \,d\theta \\[6pt] &= \int(a\cos\theta)(a\cos\theta) \,d\theta \\[6pt] &= a^2\int\cos^2\theta\,d\theta \\[6pt] &= a^2\int\left(\frac{1+\cos 2\theta}{2}\right)\,d\theta \\[6pt] &= \frac{a^2}{2} \left(\theta+\frac{1}{2}\sin 2\theta \right) + C \\[6pt] &= \frac{a^2}{2}(\theta+\sin\theta\cos\theta) + C \\[6pt] &= \frac{a^2}{2}\left(\arcsin\frac{x}{a}+\frac{x}{a}\sqrt{1-\frac{x^2}{a^2}}\right) + C \\[6pt] &= \frac{a^2}{2}\arcsin\frac{x}{a}+\frac{x}{2}\sqrt{a^2-x^2}+C. \end{align} }[/math]

For a definite integral, the bounds change once the substitution is performed and are determined using the equation [math]\displaystyle{ \theta = \arcsin\frac{x}{a}, }[/math] with values in the range [math]\displaystyle{ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}. }[/math] Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

[math]\displaystyle{ \int_{-1}^1\sqrt{4-x^2}\,dx, }[/math]

may be evaluated by substituting [math]\displaystyle{ x = 2\sin\theta, \,dx = 2\cos\theta\,d\theta, }[/math] with the bounds determined using [math]\displaystyle{ \theta = \arcsin\frac{x}{2}. }[/math]

Because [math]\displaystyle{ \arcsin(1/{2}) = \pi/6 }[/math] and [math]\displaystyle{ \arcsin(-1/2) = -\pi/6, }[/math] [math]\displaystyle{ \begin{align} \int_{-1}^1\sqrt{4-x^2}\,dx &= \int_{-\pi/6}^{\pi/6}\sqrt{4-4\sin^2\theta}\,(2\cos\theta) \,d\theta \\[6pt] &= \int_{-\pi/6}^{\pi/6}\sqrt{4(1-\sin^2\theta)}\,(2\cos\theta) \,d\theta \\[6pt] &= \int_{-\pi/6}^{\pi/6}\sqrt{4(\cos^2\theta)}\,(2\cos\theta) \,d\theta \\[6pt] &= \int_{-\pi/6}^{\pi/6}(2\cos\theta)(2\cos\theta) \,d\theta \\[6pt] &= 4\int_{-\pi/6}^{\pi/6}\cos^2\theta\,d\theta \\[6pt] &= 4\int_{-\pi/6}^{\pi/6}\left(\frac{1+\cos 2\theta}{2}\right)\,d\theta \\[6pt] &= 2 \left[\theta+\frac{1}{2} \sin 2\theta \right]^{\pi/6}_{-\pi/6} = [2\theta+\sin 2\theta] \Biggl |^{\pi/6}_{-\pi/6} \\[6pt] &= \left(\frac{\pi}{3}+\sin\frac{\pi}{3}\right)-\left(-\frac{\pi}{3}+\sin\left(-\frac{\pi}{3}\right)\right) = \frac{2\pi}{3}+\sqrt{3}. \end{align} }[/math]

On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields [math]\displaystyle{ \begin{align} \int_{-1}^1\sqrt{4-x^2}\,dx &= \left[ \frac{2^2}{2}\arcsin\frac{x}{2}+\frac{x}{2}\sqrt{2^2-x^2} \right]_{-1}^{1}\\[6pt] &= \left( 2 \arcsin \frac{1}{2} + \frac{1}{2}\sqrt{4-1}\right) - \left( 2 \arcsin \left(-\frac{1}{2}\right) + \frac{-1}{2}\sqrt{4-1}\right)\\[6pt] &= \left( 2 \cdot \frac{\pi}{6} + \frac{\sqrt{3}}{2}\right) - \left( 2\cdot \left(-\frac{\pi}{6}\right) - \frac{\sqrt 3}{2}\right)\\[6pt] &= \frac{2\pi}{3} + \sqrt{3} \end{align} }[/math] as before.

Case II: Integrands containing a² + x²

Let [math]\displaystyle{ x = a \tan \theta, }[/math] and use the identity [math]\displaystyle{ 1+\tan^2 \theta = \sec^2 \theta. }[/math]

Examples of Case II

Geometric construction for Case II

Example 1

In the integral

[math]\displaystyle{ \int\frac{dx}{a^2+x^2} }[/math]

we may write

[math]\displaystyle{ x=a\tan\theta,\quad dx=a\sec^2\theta\, d\theta, \quad \theta=\arctan\frac{x}{a}, }[/math]

so that the integral becomes

[math]\displaystyle{ \begin{align} \int\frac{dx}{a^2+x^2} &= \int\frac{a\sec^2\theta\, d\theta}{a^2 + a^2\tan^2\theta} \\[6pt] &= \int\frac{a\sec^2\theta\, d\theta}{a^2(1+\tan^2\theta)} \\[6pt] &= \int\frac{a\sec^2\theta\, d\theta}{a^2\sec^2\theta} \\[6pt] &= \int\frac{d\theta}{a} \\[6pt] &= \frac{\theta}{a}+C \\[6pt] &= \frac{1}{a} \arctan \frac{x}{a} + C, \end{align} }[/math]

provided [math]\displaystyle{ a \neq 0. }[/math]

For a definite integral, the bounds change once the substitution is performed and are determined using the equation [math]\displaystyle{ \theta = \arctan\frac{x}{a}, }[/math] with values in the range [math]\displaystyle{ -\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}. }[/math] Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

[math]\displaystyle{ \int_0^1\frac{4\, dx}{1+x^2}\, }[/math]

may be evaluated by substituting [math]\displaystyle{ x = \tan\theta, \,dx = \sec^2\theta\,d\theta, }[/math] with the bounds determined using [math]\displaystyle{ \theta = \arctan x. }[/math]

Since [math]\displaystyle{ \arctan 0 = 0 }[/math] and [math]\displaystyle{ \arctan 1 = \pi/4, }[/math] [math]\displaystyle{ \begin{align} \int_0^1\frac{4\,dx}{1+x^2} &= 4\int_0^1\frac{dx}{1 + x^2} \\[6pt] &= 4\int_0^{\pi/4}\frac{\sec^2\theta\, d\theta}{1+\tan^2\theta} \\[6pt] &= 4\int_0^{\pi/4}\frac{\sec^2\theta\, d\theta}{\sec^2\theta} \\[6pt] &= 4\int_0^{\pi/4}d\theta \\[6pt] &= (4\theta)\Bigg|^{\pi/4}_0 = 4 \left (\frac{\pi}{4} - 0 \right) = \pi. \end{align} }[/math]

Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields [math]\displaystyle{ \begin{align} \int_0^1\frac{4\,dx}{1+x^2}\, &= 4\int_0^1\frac{dx}{1+x^2} \\ &= 4\left[\frac{1}{1} \arctan \frac{x}{1} \right]^1_0 \\ &= 4(\arctan x)\Bigg|^1_0 \\ &= 4(\arctan 1 - \arctan 0) \\ &= 4 \left (\frac{\pi}{4} - 0 \right) = \pi, \end{align} }[/math] same as before.

Example 2

The integral

[math]\displaystyle{ \int\sqrt{a^2+x^2}\,{dx} }[/math]

may be evaluated by letting [math]\displaystyle{ x=a\tan\theta,\, dx=a\sec^2\theta\, d\theta, \, \theta=\arctan\frac{x}{a}, }[/math]

where [math]\displaystyle{ a \gt 0 }[/math] so that [math]\displaystyle{ \sqrt{a^2}=a, }[/math] and [math]\displaystyle{ -\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2} }[/math] by the range of arctangent, so that [math]\displaystyle{ \sec \theta \gt 0 }[/math] and [math]\displaystyle{ \sqrt{\sec^2 \theta} = \sec \theta. }[/math]

Then, [math]\displaystyle{ \begin{align} \int\sqrt{a^2+x^2}\,dx &= \int\sqrt{a^2 + a^2\tan^2\theta}\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int\sqrt{a^2 (1+\tan^2\theta)}\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int\sqrt{a^2 \sec^2\theta}\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int(a \sec\theta)(a \sec^2\theta)\, d\theta \\[6pt] &= a^2\int \sec^3\theta\, d\theta. \\[6pt] \end{align} }[/math] The integral of secant cubed may be evaluated using integration by parts. As a result, [math]\displaystyle{ \begin{align} \int\sqrt{a^2+x^2}\,dx &= \frac{a^2}{2}(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)+C \\[6pt] &= \frac{a^2}{2}\left(\sqrt{1+\frac{x^2}{a^2}}\cdot\frac{x}{a} + \ln\left|\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right|\right)+C \\[6pt] &= \frac{1}{2}\left(x\sqrt{a^2+x^2} + a^2\ln\left|\frac{x+\sqrt{a^2+x^2}}{a}\right|\right)+C. \end{align} }[/math]

Case III: Integrands containing x² − a²

Let [math]\displaystyle{ x = a \sec \theta, }[/math] and use the identity [math]\displaystyle{ \sec^2 \theta -1 = \tan^2 \theta. }[/math]

Examples of Case III

Geometric construction for Case III

Integrals like

[math]\displaystyle{ \int\frac{dx}{x^2 - a^2} }[/math]

can also be evaluated by partial fractions rather than trigonometric substitutions. However, the integral

[math]\displaystyle{ \int\sqrt{x^2 - a^2}\, dx }[/math]

cannot. In this case, an appropriate substitution is: [math]\displaystyle{ x = a \sec\theta,\, dx = a \sec\theta\tan\theta\, d\theta, \, \theta = \arcsec\frac{x}{a}, }[/math]

where [math]\displaystyle{ a \gt 0 }[/math] so that [math]\displaystyle{ \sqrt{a^2}=a, }[/math] and [math]\displaystyle{ 0 \le \theta \lt \frac{\pi}{2} }[/math] by assuming [math]\displaystyle{ x \gt 0, }[/math] so that [math]\displaystyle{ \tan \theta \ge 0 }[/math] and [math]\displaystyle{ \sqrt{\tan^2 \theta} = \tan \theta. }[/math]

Then, [math]\displaystyle{ \begin{align} \int\sqrt{x^2 - a^2}\, dx &= \int\sqrt{a^2 \sec^2\theta - a^2} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int\sqrt{a^2 (\sec^2\theta - 1)} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int\sqrt{a^2 \tan^2\theta} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int a^2 \sec\theta\tan^2\theta\, d\theta \\ &= a^2 \int (\sec\theta)(\sec^2\theta - 1)\, d\theta \\ &= a^2 \int (\sec^3\theta - \sec\theta)\, d\theta. \end{align} }[/math]

One may evaluate the integral of the secant function by multiplying the numerator and denominator by [math]\displaystyle{ ( \sec \theta + \tan \theta) }[/math] and the integral of secant cubed by parts. As a result, [math]\displaystyle{ \begin{align} \int\sqrt{x^2-a^2}\,dx &= \frac{a^2}{2}(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)-a^2\ln|\sec\theta+\tan\theta|+C \\[6pt] &= \frac{a^2}{2}(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|)+C \\[6pt] &= \frac{a^2}{2}\left(\frac{x}{a}\cdot\sqrt{\frac{x^2}{a^2}-1} - \ln\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|\right)+C \\[6pt] &= \frac{1}{2}\left(x\sqrt{x^2-a^2} - a^2\ln\left|\frac{x+\sqrt{x^2-a^2}}{a}\right|\right)+C. \end{align} }[/math]

When [math]\displaystyle{ \frac{\pi}{2} \lt \theta \le \pi, }[/math] which happens when [math]\displaystyle{ x \lt 0 }[/math] given the range of arcsecant, [math]\displaystyle{ \tan \theta \le 0, }[/math] meaning [math]\displaystyle{ \sqrt{\tan^2 \theta} = -\tan \theta }[/math] instead in that case.

Substitutions that eliminate trigonometric functions

Substitution can be used to remove trigonometric functions.

For instance,

[math]\displaystyle{ \begin{align} \int f(\sin(x), \cos(x))\, dx &=\int\frac1{\pm\sqrt{1-u^2}} f\left(u,\pm\sqrt{1-u^2}\right)\, du && u=\sin (x) \\[6pt] \int f(\sin(x), \cos(x))\, dx &=\int\frac{1}{\mp\sqrt{1-u^2}} f\left(\pm\sqrt{1-u^2},u\right)\, du && u=\cos (x) \\[6pt] \int f(\sin(x), \cos(x))\, dx &=\int\frac2{1+u^2} f \left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\, du && u=\tan\left (\tfrac{x}{2} \right ) \\[6pt] \end{align} }[/math]

The last substitution is known as the Weierstrass substitution, which makes use of tangent half-angle formulas.

For example,

[math]\displaystyle{ \begin{align} \int\frac{4 \cos x}{(1+\cos x)^3}\, dx &= \int\frac2{1+u^2}\frac{4\left(\frac{1-u^2}{1+u^2}\right)}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\, du = \int (1-u^2)(1+u^2)\, du \\&= \int (1-u^4)\,du = u - \frac{u^5}{5} + C = \tan \frac{x}{2} - \frac{1}{5} \tan^5 \frac{x}{2} + C. \end{align} }[/math]

Hyperbolic substitution

Substitutions of hyperbolic functions can also be used to simplify integrals.

In the integral [math]\displaystyle{ \int \frac{dx}{\sqrt{a^2+x^2}}\,, }[/math] make the substitution [math]\displaystyle{ x=a\sinh{u}, }[/math] [math]\displaystyle{ dx=a\cosh u\, du. }[/math]

Then, using the identities [math]\displaystyle{ \cosh^2 (x) - \sinh^2 (x) = 1 }[/math] and [math]\displaystyle{ \sinh^{-1}{x} = \ln(x + \sqrt{x^2 + 1}), }[/math]

[math]\displaystyle{ \begin{align} \int \frac{dx}{\sqrt{a^2+x^2}}\, &= \int \frac{a\cosh u\,du}{\sqrt{a^2+a^2\sinh^2 u}}\ , \\[6pt] &=\int \frac{a\cosh{u}\, du}{a\sqrt{1+\sinh^2{u}}}\, \\[6pt] &=\int \frac{a\cosh{u}}{a\cosh u}\, du\\[6pt] &=u+C\\[6pt] &=\sinh^{-1}{\frac{x}{a}}+C\\[6pt] &=\ln\left(\sqrt{\frac{x^2}{a^2} + 1} + \frac{x}{a}\right) + C\\[6pt] &=\ln\left(\frac{\sqrt{x^2+a^2} + x}{a}\right) + C \end{align} }[/math]

See also

• Integration by substitution
• Weierstrass substitution
• Euler substitution