Calculus II: Trigonometric Substitution

Trigonometric substitution is a powerful technique for evaluating integrals containing radical expressions, particularly those involving forms like $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, and $\sqrt{x^2-a^2}$. Mastering this method is essential for engineering and physics applications, from calculating moments of inertia to analyzing waveforms, as it transforms an algebraic impasse into a trigonometric integral that is often much simpler to solve. The core idea is to use a trigonometric identity to eliminate the radical, turning a difficult integration problem into one involving standard trigonometric integrals.

The Three Standard Substitutions and Their Geometry

The method is built upon three primary substitutions, each designed to exploit a specific Pythagorean identity. The choice of substitution is dictated by the form of the radical present in the integrand.

1. For Integrands Containing $\sqrt{a^2-x^2}$ Use the substitution $x=a\sin \theta$. The differential is $dx=a\cos \theta d\theta$. The radical simplifies using the identity $1-{\sin }^2\theta ={\cos }^2\theta$: $$\sqrt{a^2-x^2}=\sqrt{a^2-(a\sin \theta {)}^2}=\sqrt{a^2(1-{\sin }^2\theta )}=\sqrt{a^2{\cos }^2\theta }=a|\cos \theta |.$$ Within the domain of the inverse sine function ($-\pi /2\le \theta \le \pi /2$), $\cos \theta \ge 0$, so the absolute value simplifies to $a\cos \theta$.

Example: Evaluate $\int \sqrt{9-x^2}dx$. Let $x=3\sin \theta$, so $dx=3\cos \theta d\theta$. The integral becomes: $$\int \sqrt{9-(3\sin \theta {)}^2}\cdot 3\cos \theta d\theta =\int 3\cos \theta \cdot 3\cos \theta d\theta =9\int {\cos }^2\theta d\theta .$$ Using the power-reduction identity ${\cos }^2\theta =\frac{1}{2}(1+\cos 2\theta )$, we integrate: $$9\int \frac{1}{2}(1+\cos 2\theta )d\theta =\frac{9}{2}\left(\theta +\frac{1}{2}\sin 2\theta \right)+C=\frac{9}{2}\theta +\frac{9}{4}\sin 2\theta +C.$$ We now must back-substitute to return to the variable $x$. Since $x=3\sin \theta$, we have $\theta =\arcsin (x/3)$. For $\sin 2\theta$, we use the identity $\sin 2\theta =2\sin \theta \cos \theta =2(x/3)(\sqrt{9-x^2}/3)=\frac{2x\sqrt{9-x^2}}{9}$. The final answer is: $$\int \sqrt{9-x^2}dx=\frac{9}{2}\arcsin \left(\frac{x}{3}\right)+\frac{1}{2}x\sqrt{9-x^2}+C.$$

2. For Integrands Containing $\sqrt{a^2+x^2}$ Use the substitution $x=a\tan \theta$. Then $dx=a{\sec }^2\theta d\theta$. The radical simplifies using $1+{\tan }^2\theta ={\sec }^2\theta$: $$\sqrt{a^2+x^2}=\sqrt{a^2+(a\tan \theta {)}^2}=\sqrt{a^2(1+{\tan }^2\theta )}=\sqrt{a^2{\sec }^2\theta }=a|\sec \theta |.$$ For $\theta$ in $(-\pi /2,\pi /2)$, $\sec \theta >0$, so this becomes $a\sec \theta$.

3. For Integrands Containing $\sqrt{x^2-a^2}$ Use the substitution $x=a\sec \theta$. Then $dx=a\sec \theta \tan \theta d\theta$. The radical simplifies using ${\sec }^2\theta -1={\tan }^2\theta$: $$\sqrt{x^2-a^2}=\sqrt{(a\sec \theta {)}^2-a^2}=\sqrt{a^2({\sec }^2\theta -1)}=\sqrt{a^2{\tan }^2\theta }=a|\tan \theta |.$$ The sign depends on the domain. For $x>a$ (implying $\sec \theta >1$), we typically take $\theta$ in $[0,\pi /2)$ where $\tan \theta \ge 0$, so it simplifies to $a\tan \theta$.

Back-Substitution Using Right Triangles

After integrating in terms of $\theta$, you must express the result in the original variable $x$. While inverse trigonometric functions like $\arcsin (x/a)$ are direct, terms like $\sin (2\arcsin (x/a))$ are messy. A clearer method is to use a right triangle constructed from the original substitution.

For the example $x=3\sin \theta$, we build a right triangle where $\sin \theta =\frac{x}{3}=\frac{\text{opposite}}{\text{hypotenuse}}$. Label the side opposite $\theta$ as $x$ and the hypotenuse as $3$. The adjacent side, by the Pythagorean Theorem, is $\sqrt{9-x^2}$. From this triangle, we can read off: $\sin \theta =x/3$, $\cos \theta =\sqrt{9-x^2}/3$, and $\theta =\arcsin (x/3)$. This visual approach is faster and less error-prone than manipulating double-angle identities.

Handling Integrals That Require Completing the Square First

Not every quadratic expression fits the three standard forms immediately. You may first need to complete the square to transform it into one of the recognizable patterns.

Example: Evaluate $\int \frac{dx}{\sqrt{x^2+4x+13}}$. Complete the square for the quadratic inside the radical: $x^2+4x+13=(x^2+4x+4)+9=(x+2{)}^2+3^2$. The integral becomes $\int \frac{dx}{\sqrt{(x+2{)}^2+3^2}}$, which matches the $\sqrt{a^2+u^2}$ form with $u=x+2$ and $a=3$. Use the substitution $x+2=3\tan \theta$, then $dx=3{\sec }^2\theta d\theta$, and proceed as outlined in the standard process.

Definite Integral Adjustments

When evaluating a definite integral using trigonometric substitution, you have two equivalent paths. First, you can convert the limits of integration from $x$-values to corresponding $\theta$-values using your substitution function. This often simplifies the final computation because you never need to revert to the $x$ variable. Alternatively, you can find the antiderivative in terms of $x$ and then apply the original $x$-limits. The first method is generally more efficient.

For example, to evaluate ${\int }_0^{3/2}\frac{dx}{\sqrt{9-x^2}}$ with substitution $x=3\sin \theta$:

• When $x=0$, $0=3\sin \theta \Rightarrow \theta =0$.
• When $x=3/2$, $3/2=3\sin \theta \Rightarrow \sin \theta =1/2\Rightarrow \theta =\pi /6$.

The integral transforms to ${\int }_{\theta =0}^{\theta =\pi /6}\frac{3\cos \theta d\theta }{3\cos \theta }={\int }_0^{\pi /6}d\theta =\pi /6$.

Combining Trigonometric Substitution with Other Techniques

Trigonometric substitution is rarely the only step. You will frequently need to combine it with other integration techniques. The resulting integral in $\theta$ often involves powers of sine, cosine, secant, or tangent, requiring methods like:

• Power-reduction identities (as shown in the first example).
• Integration by parts.
• Further substitution (e.g., $u=\cos \theta$ for integrals of the form $\int {\sin }^n\theta \cos \theta d\theta$).

A common workflow is: 1) Identify the radical form and choose the correct trig substitution. 2) Simplify the integral, eliminating the radical. 3) Integrate the resulting trigonometric expression using appropriate methods. 4) Back-substitute to express the antiderivative in the original variable using a right triangle or inverse trig functions.

Common Pitfalls

1. Ignoring the Absolute Value During Simplification: The most frequent algebraic error is forgetting that $\sqrt{u^2}=|u|$, not simply $u$. For example, $\sqrt{a^2{\cos }^2\theta }=a|\cos \theta |$. You must consider the domain of $\theta$ implied by your inverse substitution to correctly remove the absolute value bars. Using the standard intervals for $\theta$ (e.g., $-\pi /2\le \theta \le \pi /2$ for $x=a\sin \theta$) ensures the trigonometric function is non-negative.

1. Incorrect Differential or Misplaced Constant: When you set $x=a\sin \theta$, remember $dx=a\cos \theta d\theta$. The constant $a$ must be included in the differential. A missed constant will lead to an incorrect result. Always compute the differential immediately after stating the substitution.

1. Mishandling Definite Integral Limits: If you choose to change the limits of integration to $\theta$, you must completely convert the integral to the new variable, including the differential and the limits. Do not mix $x$ and $\theta$ in the same evaluation step. If you instead back-substitute to $x$ first, you must use the original $x$-limits.

1. Forgetting to Back-Substitute Completely: After integrating with respect to $\theta$, your final answer must be in terms of the original variable (unless you changed limits for a definite integral). It's easy to leave an answer as $\frac{9}{2}\theta +\frac{9}{2}\sin \theta \cos \theta +C$. You must use your substitution or right triangle to express $\theta$, $\sin \theta$, $\cos \theta$, etc., in terms of $x$.

Summary

• Trigonometric substitution converts integrals with radicals $\sqrt{a^2\pm x^2}$ or $\sqrt{x^2-a^2}$ into trigonometric integrals by using the substitutions $x=a\sin \theta$, $x=a\tan \theta$, and $x=a\sec \theta$, respectively.
• The process hinges on Pythagorean identities to eliminate the radical: $1-{\sin }^2\theta ={\cos }^2\theta$, $1+{\tan }^2\theta ={\sec }^2\theta$, and ${\sec }^2\theta -1={\tan }^2\theta$.
• Always account for the differential ($dx$) when making the substitution and remember that $\sqrt{u^2}=|u|$, simplifying based on the chosen domain for $\theta$.
• Back-substitution to the original variable $x$ is efficiently done by constructing a right triangle based on the original substitution equation.
• For quadratics not in standard form, completing the square is a necessary first step to reveal the appropriate structure for substitution.
• In definite integrals, you can convert the limits of integration to $\theta$-values to streamline the calculation, avoiding the need to back-substitute.