6.3: Trigonometric Substitution

In Section 4.4 we studied a fundamental technique for evaluating integrals—by substitution. We now expand upon this idea by considering unique substitutions, trigonometric substitutions, which allow us to integrate functions containing radicals. Some examples include \[\andThree{\int \sqrt{100 - x^2} \di x}{\int \frac{x^2 - 1}{x \sqrt{x^2 + 16}} \di x}{\int \frac{3x - 6}{\sqrt{4x^2 - 100}} \di x} \pd\] In this section we learn to convert these integrals into trigonometric integrals.

First consider the family of integrals \(\int x \sqrt{a^2 - x^2} \di x,\) \(a \ne 0.\) To evaluate it, we substitute \(u = a^2 - x^2\) because then \(\dd u = -2x \di x,\) effectively stripping the extra \(x\) in the integrand. But evaluating \(\int \sqrt{a^2 - x^2} \di x,\) \(a \gt 0,\) requires a special plan—we first remove the root sign. Let's perform the trigonometric substitution \(x = a \sin \theta.\) Then, by the Pythagorean identity in the form \(1 - \sin^2 \theta = \cos^2 \theta,\) \[ \ba \sqrt{a^2 - x^2} &= \sqrt{a^2 - (a \sin \theta)^2} \nl &= a \sqrt{1 - \sin^2 \theta} \nl &= a \sqrt{\cos^2 \theta} \nl &= a \abs{\cos \theta} \pd \ea \] Integrating \(a \cos \theta\) is far easier than integrating \(\sqrt{a^2 - x^2}.\) Thus, stripping the root sign will become our primary goal.

Next we examine two other radical forms. For \(\sqrt{x^2 - a^2},\) we try \(x = a \sec \theta.\) Then since \(\sec^2 \theta - 1 = \tan^2 \theta,\) we have \[ \ba \sqrt{x^2 - a^2} &= \sqrt{(a \sec \theta)^2 - a^2} \nl &= a \sqrt{\sec^2 \theta - 1} \nl &= a \sqrt{\tan^2 \theta} \nl &= a \abs{\tan \theta} \pd \ea \] Similarly, for the case \(\sqrt{x^2 + a^2}\) we substitute \(x = a \tan \theta.\) Because \(\tan^2 \theta + 1 = \sec^2 \theta,\) \[ \ba \sqrt{x^2 + a^2} &= \sqrt{(a \tan \theta)^2 + a^2} \nl &= a \sqrt{\tan^2 \theta + 1} \nl &= a \sqrt{\sec^2 \theta} \nl &= a \abs{\sec \theta} \pd \ea \] The following table summarizes the substitutions for all three cases. In each case, we restrict the values of \(\theta\) so that the substitution forms a function that is one-to-one on the selected interval.

Now we combine trigonometric substitutions with the Substitution Rule to evaluate indefinite integrals. After choosing a trigonometric substitution, we compute the differential. For example, if we choose \(x = 3 \sin \theta,\) then the differential is \(\dd x = 3 \cos \theta \di \theta.\) Then we simply substitute these expressions to convert the integral in terms of \(\theta.\) The following steps summarize this process.

The denominator has the factor \(\sqrt{4 - x^2},\) so we choose the trigonometric substitution \(x = 2 \sin \theta\) \((-\pi/2 \leq \theta \leq \pi/2).\) Then \(\dd x = 2 \cos \theta \di \theta,\) and the integral becomes \[ \int \frac{1}{(2 \sin \theta) \sqrt{4 - 4 \sin^2 \theta}} (2 \cos \theta) \di \theta = \int \frac{\cos \theta}{2 \sin \theta \abs{\cos \theta}} \di \theta \pd \] Note that \(\abs{\cos \theta}\) \(= \cos \theta\) on \([-\pi/2, \pi/2],\) enabling us to drop the absolute value bars. Hence, \[ \ba \int \frac{\cos \theta}{2 \sin \theta \abs{\cos \theta}} \di \theta &= \tfrac{1}{2} \int \csc \theta \di \theta \nl &= -\tfrac{1}{2} \ln \abs{\csc \theta + \cot \theta} + C \pd \ea \] Now we convert our answer back to \(x.\) Since \(x = 2 \sin \theta,\) we have \(\sin \theta = x/2.\) Accordingly, in a reference right triangle of hypotenuse \(2,\) \(x\) is the side opposite \(\theta.\) Then the side adjacent to \(\theta\) is, by the Pythagorean Theorem, \(\sqrt{2^2 - x^2}\) \(= \sqrt{4 - x^2}.\) (See Figure 1.) Through trigonometric relationships, we see \[\csc \theta = \frac{2}{x} \and \cot \theta = \frac{\sqrt{4 - x^2}}{x} \pd\] Although \(\theta \gt 0\) in the diagram, these expressions are valid even for \(\theta \lt 0.\) Thus, \[\int \frac{1}{x \sqrt{4 - x^2}} \di x = \boxed{-\frac{1}{2} \ln \abs{\frac{2}{x} + \frac{\sqrt{4 - x^2}}{x}} + C}\]

The expression \(\sqrt{a^2 + x^2}\) warrants the trigonometric substitution \(x = a \tan \theta,\) \(-\pi/2 \lt \theta \lt \pi/2.\) Then \(\dd x = a \sec^2 \theta \di \theta,\) and the integral becomes \[ \ba \int \frac{1}{\sqrt{a^2 + a^2 \tan^2 \theta}} (a \sec^2 \theta) \di \theta &= \int \frac{a \sec^2 \theta}{\sqrt{a^2 \sec^2 \theta}} \di \theta \nl &= \int \frac{\sec^2 \theta}{\abs{\sec \theta}} \di \theta \pd \ea \] On \((-\pi/2, \pi/2),\) secant is positive and so \(\abs{\sec \theta} = \sec \theta.\) The integral therefore becomes \[\int \sec \theta \di \theta = \ln \abs{\sec \theta + \tan \theta} + C_1 \pd \] Now we rewrite this answer in terms of \(x.\) Because \(x = a \tan \theta,\) it follows that \(\tan \theta = x/a.\) Thus, we construct a reference right triangle containing an angle \(\theta\) whose opposite side has length \(x\) and adjacent side has length \(a.\) By the Pythagorean Theorem, the hypotenuse has length \(\sqrt{a^2 + x^2}.\) (See Figure 2.) We then see \(\sec \theta = \sqrt{a^2 + x^2}/a.\) Therefore, the integral becomes \[\ln \abs{\frac{\sqrt{a^2 + x^2}}{a} + \frac{x}{a}} + C_1 \pd \] Using logarithm laws and noting that \(a \gt 0,\) we have \[\ln \abs{\sqrt{a^2 + x^2} + x} - \ln a + C_1 \pd\] Since \(a\) is a constant, we denote \(C_1 - \ln a\) as one constant: \(C.\) Our final answer is then, as requested, \[\ln \abs{\sqrt{a^2 + x^2} + x} + C \pd \]

Now we examine performing trigonometric substitutions to evaluate definite integrals. In these problems, it is easiest to convert the bounds to be in terms of \(\theta,\) as shown by the following examples. Remember not to omit any absolute value bars; we resolve them based on the function's behavior between the integral's bounds.

Solving for \(y\) as a function of \(x\) yields \[y = \pm b \sqrt{1 - \frac{x^2}{a^2}} \pd\] The top portion of the ellipse is given by the positive solution, so the area in the first quadrant (as shown in Figure 3) is \[ \ba A_1 &= \int_0^a b \sqrt{1 - \frac{x^2}{a^2}} \di x \nl &= \frac{b}{a} \int_0^a \sqrt{a^2 - x^2} \di x \pd \ea \] To evaluate this integral, we substitute \(x = a \sin \theta,\) \(-\pi/2 \leq \theta \leq \pi/2.\) Then \(\dd x = a \cos \theta \di \theta.\) We now change the bounds of integration as follows.

• Lower Bound When \(x = 0,\) we have \(a \sin \theta = 0\) or \(\sin \theta = 0.\) Thus, \(\theta = 0.\)
• Upper Bound When \(x = a,\) it follows that \(a \sin \theta = a\) or \(\sin \theta = 1.\) The only solution in \([-\pi/2, \pi/2]\) is \(\theta = \pi/2.\)

We see \[ \ba \sqrt{a^2 - x^2} &= \sqrt{a^2 - a^2 \sin^2 \theta} \nl &= \sqrt{a^2 \cos^2 \theta} \nl &= a \abs{\cos \theta} \nl &= a \cos \theta \cma \ea \] where the last step is true because cosine is nonnegative on \([0, \pi/2].\) Accordingly, we have \[ \ba A_1 &= \frac{b}{a} \int_0^{\pi/2} a \cos \theta \, (a \cos \theta) \di \theta \nl &= ab \int_0^{\pi/2} \cos^2 \theta \di \theta \nl &= ab \int_0^{\pi/2} \tfrac{1}{2} (1 + \cos 2 \theta) \di \theta \nl &= ab \par{\tfrac{1}{2} \theta + \tfrac{1}{4} \sin 2 \theta} \intEval_0^{\pi/2} \nl &= \frac{\pi}{4} ab \pd \ea \] Since the ellipse is symmetric with respect to both axes, the total area is four times the area in the first quadrant—namely, \[A = 4A_1 = \boxed{\pi ab}\] If \(a = b = r,\) then the ellipse becomes a circle of radius \(r;\) indeed, the area formula becomes the famous \(A = \pi r^2.\)

In the following examples, we solve more complicated integrals requiring trigonometric substitutions. See Section 6.2 to review the strategies for resolving trigonometric integrals.

We complete the square under the radical to get \[\int \frac{3x}{\sqrt{(x - 1)^2 + 4}} \di x \pd\] We let \(x - 1 = 2 \tan \theta,\) where \(-\pi/2 \lt \theta \lt \pi/2.\) Then \[ \ba x &= 1 + 2 \tan \theta \cma \nl \dd x &= 2 \sec^2 \theta \di \theta \pd \ea \] Afterward, the integral becomes \[ \int \frac{3(1 + 2 \tan \theta)}{\sqrt{4 \tan^2 \theta + 4}} (2 \sec^2 \theta) \di \theta = 3 \int \frac{1 + 2 \tan \theta}{\abs{\sec \theta}} \, \sec^2 \theta \di \theta \pd \] Because secant is positive on \((-\pi/2, \pi/2),\) we have \(\abs{\sec \theta}\) \(= \sec \theta\) and so the integral becomes \[ \ba 3 \int \frac{1 + 2 \tan \theta}{\sec \theta} \, \sec^2 \theta \di \theta &= 3 \int \par{\sec \theta + 2 \sec \theta \tan \theta} \di \theta \nl &= 3 \ln \abs{\sec \theta + \tan \theta} + 6 \sec \theta + C_1 \pd \ea \] Lastly, we express the answer in terms of \(x.\) We substituted \(x - 1 = 2 \tan \theta,\) so it follows that \(\tan \theta = (x - 1)/2.\) Thus, we draw a reference right triangle with legs of lengths \((x - 1)\) and \(2,\) which create a hypotenuse of length \(\sqrt{4 + (x - 1)^2}.\) (See Figure 4.) We then see \[\sec \theta = \frac{\sqrt{4 + (x - 1)^2}}{2} \cma\] so we have \[ 3 \ln \abs{\frac{\sqrt{4 + (x - 1)^2}}{2} + \frac{x - 1}{2}} + 6 \par{\frac{\sqrt{4 + (x - 1)^2}}{2}} + C_1 \pd \] Simplifying then yields \[ 3 \ln \abs{\frac{\sqrt{x^2 - 2x + 5}}{2} + \frac{x - 1}{2}} + 3 \sqrt{x^2 - 2x + 5} + C_1 \pd \] By logarithm laws, this expression is equivalent to \[3 \ln \abs{\sqrt{x^2 - 2x + 5} + x - 1} - 3 \ln 2 + 3 \sqrt{x^2 - 2x + 5} + C_1 \pd\] It is easiest to combine \(-3 \ln 2 + C_1\) into simply one constant: \(C.\) Doing so yields \[\boxed{3 \ln \abs{\sqrt{x^2 - 2x + 5} + x - 1} + 3 \sqrt{x^2 - 2x + 5} + C}\]

We rewrite the denominator as \[\par{\sqrt{9x^2 + 25}}^3 = \par{3 \sqrt{x^2 + \tfrac{25}{9}}}^3 = 27 \par{\sqrt{x^2 + \tfrac{25}{9}}}^3 \pd\] This form reveals the need for the substitution \(x = \tfrac{5}{3} \tan \theta,\) \(-\pi/2 \lt \theta \lt \pi/2.\) Then \(\dd x = \frac{5}{3} \sec^2 \theta \di \theta,\) and the denominator becomes \[ \ba 27 \par{\sqrt{\par{\tfrac{5}{3} \tan \theta}^2 + \tfrac{25}{9}}}^3 &= 27 \par{\sqrt{\tfrac{25}{9} \sec^2 \theta}}^3 \nl &= 125 \, \abs{\sec \theta}^3 \pd \ea \] The integral therefore becomes \[ \int \frac{\par{\frac{5}{3} \tan \theta}^3}{125 \, \abs{\sec \theta}^3} \par{\tfrac{5}{3} \sec^2 \theta} \di \theta = \frac{5}{81} \int \frac{\tan^3 \theta}{\abs{\sec \theta}^3} \, \sec^2 \theta \di \theta \pd \] But secant is positive on \((-\pi/2, \pi/2),\) so we have \(\abs{\sec \theta}\) \(= \sec \theta.\) Hence, the integral becomes \[ \frac{5}{81} \int \frac{\tan^3 \theta}{\sec^3 \theta} \, \sec^2 \theta \di \theta = \frac{5}{81} \int \frac{\tan^3 \theta}{\sec \theta} \di \theta \pd \] It is a good idea to split the fraction of complicated trigonometric functions into sines and cosines; doing so gives \[ \ba \frac{5}{81} \int \frac{\sin^3 \theta}{\cos^2 \theta} \di \theta &= \frac{5}{81} \int \frac{\sin^2 \theta \sin \theta}{\cos^2 \theta} \di \theta \nl &= \frac{5}{81} \int \frac{(1 - \cos^2 \theta) \sin \theta}{\cos^2 \theta} \di \theta \pd \ea \] The integrand contains an extra factor \(\sin \theta,\) which we strip by substituting \(u = \cos \theta.\) Consequently, \(\dd u = - \sin \theta \di \theta\) and so the integral becomes \[ \ba -\frac{5}{81} \int \frac{1 - u^2}{u^2} \di u &= \frac{5}{81} \int \par{1 - \frac{1}{u^2}} \di u \nl &= \frac{5}{81} \par{u + \frac{1}{u}} + C \nl &= \frac{5}{81} \par{\cos \theta + \frac{1}{\cos \theta}} + C \nl &= \frac{5}{81} \par{\cos \theta + \sec \theta} + C \pd \ea \] Finally, we convert back to \(x.\) Since \(x = \tfrac{5}{3} \tan \theta,\) it follows that \(\tan \theta = 3x/5.\) Therefore, we construct a reference right triangle and label an angle \(\theta\) whose opposite side has length \(3x\) and adjacent side has length \(5.\) By the Pythagorean Theorem, the hypotenuse has length \(\sqrt{(3x)^2 + 5^2} = \sqrt{9x^2 + 25}.\) (See Figure 5.) Thus, we see \[\cos \theta = \frac{5}{\sqrt{9x^2 + 25}} \and \sec \theta = \frac{\sqrt{9x^2 + 25}}{5} \pd\] So our final answer is \[\frac{5}{81} \par{\frac{5}{\sqrt{9x^2 + 25}} + \frac{\sqrt{9x^2 + 25}}{5}} + C = \boxed{\frac{25}{81 \sqrt{9x^2 + 25}} + \frac{\sqrt{9x^2 + 25}}{81} + C} \]