Trigonometric Substitutions

Section 3.3 Trigonometric Substitutions

In this section we discuss how to compute indefinite integrals involving the square root of a quadratic polynomial or its reciprocal. The idea is to convert the integrand, via a suitable substitution, into a trigonometric function which can be handled by the techniques discussed in Section 3.2. This idea works for quadratic polynomials because the Pythagorean theorem imposes a quadratic relation on the sides of a right triangle.

By factoring out the leading coefficient and completing the square, we can assume the quadratic polynomial involved has leading coefficient is 1 and has no linear term. Over the real numbers, the square root of such a quadratic polynomial is in one of the following three forms: (i)\(\sqrt{a^2-x^2}, (ii) \sqrt{x^2-a^2}\) or (iii) \(\sqrt{x^2+a^2}\ (a > 0)\text{.}\) We associate to each case a right-triangle that reminds us of the corresponding substitution.

Figure 3.25. \(u = a\sin(\theta),\ -\pi/2 \le \theta \le \pi/2\)

Figure 3.26. \(u = a\sec(\theta)\text{,}\) \(0\le \theta \le \pi\) and \(\theta \neq \pi/2\text{.}\)

Figure 3.27. \(u = a\tan(\theta),\ -\pi/2 \lt \theta \lt \pi/2\)

It should be pointed out that the choice of the trigonometric functions is not unique. For instance, the substitution \(u=a\cot(\theta)\) works as well as the substitution \(u=a\tan(\theta)\) in dealing with an integral involving \(\sqrt{a^2+u^2}\text{.}\)

The substitution \(u = a\sin(\theta)\) turns \(\sqrt{a^2-u^2}\text{,}\) into

\begin{equation*} a|\cos(\theta)| = a\cos(\theta)\quad (-\pi/2 \le \theta \le \pi/2). \end{equation*}

The substitution \(u = a\tan(\theta)\) turns \(\sqrt{a^2 + u^2}\) into

\begin{equation*} a|\sec(\theta)| = a\sec(\theta)\quad (-\pi/2 \lt \theta \lt \pi/2) \end{equation*}

and the substitution \(u=a\sec(\theta)\) turns \(\sqrt{u^2-a^2}\) into \(a|\tan(\theta)|\text{.}\) However, note that

\begin{align*} |\tan(\theta)| = \begin{cases} \phantom{-}\tan(\theta) & 0 \le \theta \lt \pi/2; \\ -\tan(\theta) & \pi/2 \lt \theta \le \pi. \end{cases} \end{align*}

Example 3.28.

Let us compute \(\ds \int \frac{dx}{(x^2+9)^2}\text{.}\) The discussion above suggests the substitution \(x = 3\tan \theta\text{.}\) So, \(dx = 3\sec^2(\theta)d\theta\) and \(x^2 + 9 = 9\sec^2(\theta)\text{.}\) Thus,

\begin{align*} \int \frac{dx}{(x^2+9)^2} &= \int \frac{3\sec^2\theta}{(9\sec^2\theta)^2} d\theta = \frac{1}{27}\int \cos^2(\theta)d\theta. \end{align*}

The last integral can be handled by techniques in Section 3.2

\begin{align*} \int \cos^2(\theta)d\theta &= \int \frac{\cos(2\theta) +1}{2} d\theta \\ &= \frac{1}{2}\left( \frac{1}{2}\sin(2\theta) + \theta \right) +C\\ &= \frac{1}{2}\left( \sin\theta\cos\theta + \theta \right) + C. \end{align*}

From the corresponding diagram (Figure 3.27 with \(a=3\)) we get \(\theta = \arctan(x/3)\) and that

\begin{equation*} \sin(\theta)\cos(\theta) = \frac{x}{\sqrt{x^2+9}}\frac{3}{\sqrt{x^2+9}} = \frac{3x}{x^2+9}. \end{equation*}

Hence,

\begin{equation*} \int \frac{dx}{(x^2+9)^2} = \frac{1}{54}\left(\frac{3x}{x^2+9} + \arctan\left( \frac{x}{3} \right)\right) + C. \end{equation*}

Here is an integral involving \(\sqrt{x^2-a^2}\text{.}\)

Example 3.29.

Let us compute \(\ds \int \frac{dx}{x\sqrt{x^2-1}}\text{.}\) This time the domain of the integrand consists of two intervals. We handle the integral on each interval separately. First on \(x \gt 1\text{,}\) with the substitution \(x = \sec(\theta) (0 \lt \theta \lt \pi/2)\) suggested by the diagram Figure 3.26,

\begin{align*} \int \frac{dx}{x\sqrt{x^2-1}} & = \int \frac{d\sec(\theta)}{\sec(\theta)|\tan(\theta)|} = \int \frac{\sec(\theta)\tan(\theta)d\theta}{\sec(\theta)\tan(\theta)}\\ &= \int d\theta = \theta +C.\\ &= \asec(x) +C. (x \gt 1) \end{align*}

On \(x \lt -1\text{,}\) \(\pi/2 \lt \theta \lt \pi\) and \(|\tan(\theta)| = -\tan(\theta)\text{.}\) So, the integral equals \(-\asec(x)+C\text{.}\) Putting these together, we get

\begin{align*} \int \frac{dx}{x\sqrt{x^2-1}} &= \begin{cases} \phantom{-}\asec(x)+C & x \gt 1 \\ -\asec(x)+C & x \lt -1 \end{cases}\\ &= \arctan(\sqrt{x^2-1}) +C. \end{align*}

Recall that \(C\) represents the set of locally constant functions.

There is another substitution (Exercise 3.5.7) that works for this integral.

Figure 3.30. \(\asec(x)\) and \(\arctan(\sqrt{x^2-1})\)

Let us see an example involving \(\sqrt{a^2-x^2}\text{.}\)

Example 3.31.

To compute \(\ds \int \frac{dx}{x\sqrt{1-x^2}}\text{,}\) as Figure 3.25 suggests, let \(x = \sin(\theta) (-\pi/2 \le \theta \le \pi/2)\text{.}\) Then

\begin{align*} \int \frac{dx}{x\sqrt{1-x^2}} &= \int \frac{d\sin(\theta)}{\sin(\theta)|\cos(\theta)|}\\ & = \int \frac{\cos(\theta) d \theta}{\sin(\theta)\cos(\theta)} = \int \csc(\theta) d\theta\\ & = \ln|\csc(\theta) - \cot(\theta)| + C = \ln\left|\frac{1 - \sqrt{1-x^2}}{x}\right| + C. \end{align*}

Checkpoint 3.32.

Compute \(\ds \int \frac{dx}{\sqrt{4x^2+8x-5}}\text{.}\)

Solution.

Completing the square yields \(4x^2 + 8x -5= 4(x+1)^2 - 9\text{.}\) So, the substitution \(u=2(x+1)\) turns the integral into

\begin{equation*} \frac{1}{2} \int \frac{1}{\sqrt{u^2-9}} du. \end{equation*}

Note that the integrand \(f(u) = 1/\sqrt{u^2-9}\) is an even function of \(u\text{,}\) so if \(F(u)\) is an antiderivative of \(f(u)\) on \(u \gt 3\text{,}\) then \(-F(-u)\) would be an anti-derivative \(f(u)\) on \(u \lt 3\text{.}\) On \(u \gt 3\text{,}\) we can evaluate the integral by the substitution \(u=3\sec(\theta)\text{:}\)

\begin{align*} \frac{1}{2} \int \frac{3\sec(\theta)\tan(\theta)}{3\tan(\theta)} d\theta & = \frac{1}{2}\int\sec(\theta) d\theta\\ & = \frac{1}{2}\ln|\sec(\theta) + \tan(\theta)|+C\\ & = \frac{1}{2}\ln\left| \frac{u}{3} + \frac{\sqrt{u^2-9}}{3}\right| + C\\ & = \frac{1}{2}\ln\left| u + \sqrt{u^2-9} \right| + C. \end{align*}

Let \(F(u)\) be the anti-derivative of \(f(u)\) displayed above. Then \(\frac{F(u)-F(-u)}{2}\) will be an odd antiderivative of \(f(u)\) on the whole domain of \(f(u)\text{.}\) Note that

\begin{align*} F(u)-F(-u) & = \frac{1}{2}\left(\ln\left|u+\sqrt{u^2-9}\right| - \ln\left|-u + \sqrt{(-u)^2-9}\right|\right) \\ &= \frac{1}{2}\ln \left|\frac{u+\sqrt{u^2-9}}{u-\sqrt{u^2-9}}\right| \\ & = \frac{1}{2}\ln\left|\frac{(u+\sqrt{u^2-9})^2}{9}\right|\\ &= \ln|u+\sqrt{u^2-9}|-\ln(3). \end{align*}

Therefore, the integral can be expressed as

\begin{equation*} \frac{1}{2}\ln|u+\sqrt{u^2-9}|+C = \frac{1}{2}\ln|2(x+1) + \sqrt{4x^2+8x-5}| +C \end{equation*}

where \(C\) denotes the class of locally constant functions on the domain of the integrand.

Example 3.33.

To compute the integral

\begin{equation*} \ds \int \frac{x^{10}}{(16-x^2)^{13/2}} dx, \end{equation*}

we make the substitution \(x =4\sin \theta\text{.}\) So \(dx = 4\cos \theta d\theta\) and, from the corresponding triangle, we see that

\begin{equation*} \frac{x}{\sqrt{16-x^2}} = \tan\theta,\ \text{and}\ \frac{4}{\sqrt{16-x^2}} = \sec\theta. \end{equation*}

Thus,

\begin{align*} \int \frac{x^{10}}{(16-x^2)^{13/2}} dx & = \int \left(\frac{x}{\sqrt{16-x^2}}\right)^{10} \left(\frac{1}{\sqrt{16-x^2}}\right)^3 dx \\ &= \int (\tan^{10}\theta) \left(\frac{1}{4^3} \sec^3 \theta\right) 4\cos \theta d\theta \\ &= \frac{1}{4^2} \int \tan^{10}\theta \sec^2 \theta d\theta. \end{align*}

Now make the substitution \(w = \tan \theta\) the last integral becomes

\begin{align*} \frac{1}{4^2} \int w^{10} dw & = \frac{1}{4^2} \left(\frac{w^{11}}{11}\right) + C\\ &= \frac{1}{176}\left( \tan^{11}\theta \right) +C. \\ & = \frac{x^{11}}{176(16-x^2)^{11/2}} +C. \end{align*}

Prev Top Next