Trigonometric substitution

In mathematics, a trigonometric substitution replaces a trigonometric function for another expression. In calculus, trigonometric substitutions are a technique for evaluating integrals. In this case, an expression involving a radical function is replaced with a trigonometric one. Trigonometric identities may help simplify the answer. 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 $$x = a \sin \theta,$$ and use the identity $$1-\sin^2 \theta = \cos^2 \theta.$$

Example 1
In the integral

$$\int\frac{dx}{\sqrt{a^2-x^2}},$$

we may use

$$x=a\sin \theta,\quad dx=a\cos\theta\, d\theta, \quad \theta=\arcsin\frac{x}{a}.$$

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

The above step requires that $$a > 0$$ and $$\cos \theta > 0.$$ We can choose $$a$$ to be the principal root of $$a^2,$$ and impose the restriction $$-\pi /2 < \theta < \pi /2$$ by using the inverse sine function.

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

$$\int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}=\int_0^{\pi/6} d\theta = \frac{\pi}{6}.$$

Some care is needed when picking the bounds. Because integration above requires that $$-\pi /2 < \theta < \pi /2$$, $$\theta$$ can only go from $$0$$ to $$\pi / 6.$$ Neglecting this restriction, one might have picked $$\theta$$ to go from $$\pi$$ to $$5\pi /6,$$ 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

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

Example 2
The integral

$$\int\sqrt{a^2-x^2}\,dx,$$

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

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

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

For example, the definite integral

$$\int_{-1}^1\sqrt{4-x^2}\,dx,$$

may be evaluated by substituting $$x = 2\sin\theta, \,dx = 2\cos\theta\,d\theta,$$ with the bounds determined using $\theta = \arcsin\dfrac{x}{2}.$

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

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

Case II: Integrands containing a2 + x2
Let $$x = a \tan \theta,$$ and use the identity $$1+\tan^2 \theta = \sec^2 \theta.$$

Example 1
In the integral

$$\int\frac{dx}{a^2+x^2}$$

we may write

$$x=a\tan\theta,\quad dx=a\sec^2\theta\, d\theta, \quad \theta=\arctan\frac{x}{a},$$

so that the integral becomes

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

provided $$a \neq 0.$$

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

For example, the definite integral

$$\int_0^1\frac{4\, dx}{1+x^2}\,$$

may be evaluated by substituting $$x = \tan\theta, \,dx = \sec^2\theta\,d\theta,$$ with the bounds determined using $$\theta = \arctan x.$$

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

Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields $$\begin{align} \int_0^1\frac{4\,dx}{1+x^2}\, &= 4\int_0^1\frac{dx}{1+x^2} \\[6pt] &= 4\left[\frac{1}{1} \arctan \frac{x}{1} \right]^1_0 \\[6pt] &= 4(\arctan x)\Bigg|^1_0 \\[6pt] &= 4(\arctan 1 - \arctan 0) \\[6pt] &= 4 \left (\frac{\pi}{4} - 0 \right) = \pi, \end{align}$$ same as before.

Example 2
The integral

$$\int\sqrt{a^2+x^2}\,{dx}$$

may be evaluated by letting $$x=a\tan\theta,\, dx=a\sec^2\theta\, d\theta, \, \theta=\arctan\frac{x}{a},$$

where $$a > 0$$ so that $$\sqrt{a^2}=a,$$ and $$-\frac{\pi}{2}<\theta<\frac{\pi}{2}$$ by the range of arctangent, so that $$\sec \theta > 0$$ and $$\sqrt{\sec^2 \theta} = \sec \theta.$$

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

Case III: Integrands containing x2 − a2
Let $$x = a \sec \theta,$$ and use the identity $$\sec^2 \theta -1 = \tan^2 \theta.$$

Examples of Case III


Integrals such as

$$\int\frac{dx}{x^2 - a^2}$$

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

$$\int\sqrt{x^2 - a^2}\, dx$$

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

where $$a > 0$$ so that $$\sqrt{a^2}=a,$$ and $$0 \le \theta < \frac{\pi}{2}$$ by assuming $$x > 0,$$ so that $$\tan \theta \ge 0$$ and $$\sqrt{\tan^2 \theta} = \tan \theta.$$

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

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

When $$\frac{\pi}{2} < \theta \le \pi,$$ which happens when $$x < 0$$ given the range of arcsecant, $$\tan \theta \le 0,$$ meaning $$\sqrt{\tan^2 \theta} = -\tan \theta$$ instead in that case.

Substitutions that eliminate trigonometric functions
Substitution can be used to remove trigonometric functions.

For instance,

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

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

For example,

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

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

For example, to integrate $$1/\sqrt{a^2+x^2}$$, introduce the substitution $$x=a\sinh{u}$$ (and hence $$dx=a\cosh u \,du$$), then use the identity $\cosh^2 (x) - \sinh^2 (x) = 1$ to find:

$$\begin{align} \int \frac{dx}{\sqrt{a^2+x^2}} &= \int \frac{a\cosh u \,du}{\sqrt{a^2+a^2\sinh^2 u}} \\[6pt] &=\int \frac{\cosh{u} \,du}{\sqrt{1+\sinh^2{u}}} \\[6pt] &=\int \frac{\cosh{u}}{\cosh u} \,du \\[6pt] &=u+C \\[6pt] &=\sinh^{-1}{\frac{x}{a}} + C. \end{align}$$

If desired, this result may be further transformed using other identities, such as using the relation $\sinh^{-1}{z} = \operatorname{arsinh}{z} = \ln(z + \sqrt{z^2 + 1})$: $$\begin{align} \sinh^{-1}{\frac{x}{a}} + C &=\ln\left(\frac{x}{a} + \sqrt{\frac{x^2}{a^2} + 1}\,\right) + C \\[6pt] &=\ln\left(\frac{x + \sqrt{x^2+a^2}}{a}\,\right) + C. \end{align}$$