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 $x=a\sin \theta ,$ and use the identity $1-{\sin }^2\theta ={\cos }^2\theta .$

Examples of Case I

Geometric construction for Case I

Example 1

In the integral

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

we may use

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

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

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,

$$\underset{0}{\overset{a/2}{\int }}\frac{dx}{\sqrt{a^2-x^2}}=\underset{0}{\overset{\pi /6}{\int }}d\theta =\frac{\pi }{6}.$$

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

$$\underset{0}{\overset{a/2}{\int }}\frac{dx}{\sqrt{a^2-x^2}}=\arcsin \left(\frac{x}{a}\right){|}_0^{a/2}=\arcsin \left(\frac{1}{2}\right)-\arcsin (0)=\frac{\pi }{6}$$ as before.

Example 2

The integral

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

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

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

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

For example, the definite integral

$$\underset{-1}{\overset{1}{\int }}\sqrt{4-x^2}dx,$$

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

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

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

Case II: Integrands containing a² + x²

Let $x=a\tan \theta ,$ and use the identity $1+{\tan }^2\theta ={\sec }^2\theta .$

Examples of Case II

Geometric construction for Case II

Example 1

In the integral

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

we may write

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

so that the integral becomes

$$\begin{array}{rl}\int \frac{dx}{a^2+x^2}& =\int \frac{a{\sec }^2\theta d\theta }{a^2+a^2{\tan }^2\theta }\\ & =\int \frac{a{\sec }^2\theta d\theta }{a^2(1+{\tan }^2\theta )}\\ & =\int \frac{a{\sec }^2\theta d\theta }{a^2{\sec }^2\theta }\\ & =\int \frac{d\theta }{a}\\ & =\frac{\theta }{a}+C\\ & =\frac{1}{a}\arctan \frac{x}{a}+C,\end{array}$$

provided $a\ne 0.$

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

$$\underset{0}{\overset{1}{\int }}\frac{4dx}{1+x^2}$$

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 $\arctan 1=\pi /4,$ $$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\frac{4dx}{1+x^2}& =4\underset{0}{\overset{1}{\int }}\frac{dx}{1+x^2}\\ & =4\underset{0}{\overset{\pi /4}{\int }}\frac{{\sec }^2\theta d\theta }{1+{\tan }^2\theta }\\ & =4\underset{0}{\overset{\pi /4}{\int }}\frac{{\sec }^2\theta d\theta }{{\sec }^2\theta }\\ & =4\underset{0}{\overset{\pi /4}{\int }}d\theta \\ & =(4\theta ){|}_0^{\pi /4}=4(\frac{\pi }{4}-0)=\pi .\end{array}$$

Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields $$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\frac{4dx}{1+x^2}& =4\underset{0}{\overset{1}{\int }}\frac{dx}{1+x^2}\\ & =4{[\frac{1}{1}\arctan \frac{x}{1}]}_0^1\\ & =4(\arctan x){|}_0^1\\ & =4(\arctan 1-\arctan 0)\\ & =4(\frac{\pi }{4}-0)=\pi ,\end{array}$$ 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, $$\begin{array}{rl}\int \sqrt{a^2+x^2}dx& =\int \sqrt{a^2+a^2{\tan }^2\theta }(a{\sec }^2\theta )d\theta \\ & =\int \sqrt{a^2(1+{\tan }^2\theta )}(a{\sec }^2\theta )d\theta \\ & =\int \sqrt{a^2{\sec }^2\theta }(a{\sec }^2\theta )d\theta \\ & =\int (a\sec \theta )(a{\sec }^2\theta )d\theta \\ & =a^2\int {\sec }^3\theta d\theta .\\ \end{array}$$ The integral of secant cubed may be evaluated using integration by parts. As a result, $$\begin{array}{rl}\int \sqrt{a^2+x^2}dx& =\frac{a^2}{2}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)+C\\ & =\frac{a^2}{2}(\sqrt{1+\frac{x^2}{a^2}}\cdot \frac{x}{a}+\ln |\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}|)+C\\ & =\frac{1}{2}(x\sqrt{a^2+x^2}+a^2\ln \left|\frac{x+\sqrt{a^2+x^2}}{a}\right|)+C.\end{array}$$

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

Geometric construction for Case III

Integrals like

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

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: $$x=a\sec \theta ,dx=a\sec \theta \tan \theta d\theta ,\theta =\mathrm{arcsec}\frac{x}{a},$$

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

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, $$\begin{array}{rl}\int \sqrt{x^2-a^2}dx& =\frac{a^2}{2}(\sec \theta \tan \theta +\ln |\sec \theta +\tan \theta |)-a^2\ln |\sec \theta +\tan \theta |+C\\ & =\frac{a^2}{2}(\sec \theta \tan \theta -\ln |\sec \theta +\tan \theta |)+C\\ & =\frac{a^2}{2}(\frac{x}{a}\cdot \sqrt{\frac{x^2}{a^2}-1}-\ln |\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}|)+C\\ & =\frac{1}{2}(x\sqrt{x^2-a^2}-a^2\ln \left|\frac{x+\sqrt{x^2-a^2}}{a}\right|)+C.\end{array}$$

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,

$$\begin{array}{rlrl}\int f(\sin (x),\cos (x))dx& =\int \frac{1}{\pm \sqrt{1-u^2}}f(u,\pm \sqrt{1-u^2})du& & u=\sin (x)\\ \int f(\sin (x),\cos (x))dx& =\int \frac{1}{\mp \sqrt{1-u^2}}f(\pm \sqrt{1-u^2},u)du& & u=\cos (x)\\ \int f(\sin (x),\cos (x))dx& =\int \frac{2}{1+u^2}f(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2})du& & u=\tan \left(\frac{x}{2}\right)\\ \end{array}$$

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

For example,

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

Hyperbolic substitution

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

In the integral $\int \frac{dx}{\sqrt{a^2+x^2}},$ make the substitution $x=a\sinh u,$ $dx=a\cosh udu.$

Then, using the identities ${\cosh }^2(x)-{\sinh }^2(x)=1$ and ${\sinh }^{-1}x=\ln (x+\sqrt{x^2+1}),$

$$\begin{array}{rl}\int \frac{dx}{\sqrt{a^2+x^2}}& =\int \frac{a\cosh udu}{\sqrt{a^2+a^2{\sinh }^{2}u}},\\ & =\int \frac{a\cosh udu}{a\sqrt{1+{\sinh }^{2}u}}\\ & =\int \frac{a\cosh u}{a\cosh u}du\\ & =u+C\\ & ={\sinh }^{-1}\frac{x}{a}+C\\ & =\ln (\sqrt{\frac{x^2}{a^2}+1}+\frac{x}{a})+C\\ & =\ln \left(\frac{\sqrt{x^2+a^2}+x}{a}\right)+C\end{array}$$

See also

• Integration by substitution
• Weierstrass substitution
• Euler substitution