Trigonometric substitution

Template:Short description Template:Trigonometry In mathematics, a trigonometric substitution replaces a trigonometric function with another expression. In calculus, trigonometric substitutions are a technique for evaluating integrals. In this case, a radical expression is replaced with a trigonometric one. Trigonometric identities may help simplify the answer.

In the case of a definite integral, this substitution changes the interval of integration. Alternatively, the antiderivative of the integrand may be applied to the original interval.

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

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,

$${\int }_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}={\int }_0^{\pi /6}d\theta =\frac{\pi }{6}.$$

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

$${\int }_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}=\arcsin \left(\frac{x}{a}\right){|}_0^{a/2}=\arcsin \left(\frac{1}{2}\right)-\arcsin (0)=\frac{\pi }{6}$$ 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 $-\pi /2\le \theta \le \pi /2$ by the range of arcsine, so that $\cos \theta \ge 0$ and $\sqrt{{\cos }^2\theta }=\cos \theta .$

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

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 $-\pi /2\le \theta \le \pi /2.$ 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 $\theta =\arcsin \frac{x}{2}.$

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

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

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

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

$${\int }_0^1\frac{4dx}{1+x^2}$$

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

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

To illustrate an application of this formula, suppose we wish to calculate the arc length of the parabola $y=x^2$ from $x=0$ and $x=2$. Denoting the arc length as $s$, we get

$$s={\int }_0^2\sqrt{1+(y^{\prime }{)}^2}dx={\int }_0^2\sqrt{1+4x^2}dx=2{\int }_0^2\sqrt{\frac{1}{4}+x^2}dx.$$

In this instance, we have $a=\frac{1}{2}$, which gives

$$\begin{array}{cc}s& ={(x\sqrt{\frac{1}{4}+x^2}+\frac{1}{4}\ln \left|\frac{x+\sqrt{\frac{1}{4}+x^2}}{\frac{1}{2}}\right|)}_{x=0}^{x=2}\\ & =2\sqrt{\frac{17}{4}}+\frac{1}{4}\ln (4+2\sqrt{\frac{17}{4}})\\ & =\sqrt{17}+\frac{1}{4}\ln (4+\sqrt{17})\\ & \approx 4.64678\end{array}$$

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

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

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

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,

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

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

For example,

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

Hyperbolic substitution

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

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

$$\begin{array}{cc}\int \frac{dx}{\sqrt{a^2+x^2}}& =\int \frac{a\cosh udu}{\sqrt{a^2+a^2{\sinh }^{2}u}}\\ & =\int \frac{\cosh udu}{\sqrt{1+{\sinh }^{2}u}}\\ & =\int \frac{\cosh u}{\cosh u}du\\ & =u+C\\ & ={\sinh }^{-1}\frac{x}{a}+C.\end{array}$$

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

See also

• Weierstrass substitution
• Euler substitution

References

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