Trigonometric Substitutions

Section 3.3 Trigonometric Substitutions

This section covers indefinite integrals involving square roots of quadratic polynomials. We use substitutions to convert these into trigonometric integrals (see Section 3.2), leveraging the Pythagorean theorem’s quadratic relation between right triangle sides.

🔗

By completing the square and factoring, we reduce quadratics to forms with a leading coefficient of 1 and no linear term. The three standard forms are: (i) \(\sqrt{a^2-x^2}\text{,}\) (ii) \(\sqrt{x^2-a^2}\text{,}\) and (iii) \(\sqrt{x^2+a^2}\) (where \(a > 0\)). Each corresponds to a specific right triangle substitution.

🔗

A triangle with hypotenuse labeled by a and the side opposite to the angle theta by u, the adjacent side labeled by the square root of a square minus u square. — Figure 3.25. \(u = a\sin(\theta),\ \text{-}\pi/2 \le \theta \le \pi/2\)
🔗

A triangle with hypotenuse labeled by u and the side adjacent to the angle theta by a, the opposite side labeled by the square root of u square minus a square. — Figure 3.26. \(u = a\sec(\theta)\text{,}\) \(0\le \theta \le \pi\) and \(\theta \neq \pi/2\text{.}\)
🔗

A triangle with the side opposite to the angle theta labeled by u, the adjacent side labeled by a and the hypotenuse labeled by the square root of a square plus u square. — Figure 3.27. \(u = a\tan(\theta),\ \text{-}\pi/2 \lt \theta \lt \pi/2\)
🔗

Trigonometric choices are not unique. For example, \(u=a\cot(\theta)\) works as well as \(u=a\tan(\theta)\) for \(\sqrt{a^2+u^2}\text{.}\)

🔗

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

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

Similarly, \(u = a\tan(\theta)\) converts \(\sqrt{a^2 + u^2}\) to

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

and \(u=a\sec(\theta)\) transforms \(\sqrt{u^2-a^2}\) into \(a|\tan(\theta)|\text{.}\) 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.

To compute \(\ds \int \frac{dx}{(x^2+9)^2}\text{,}\) let \(x = 3\tan \theta\text{.}\) Then \(dx = 3\sec^2(\theta)d\theta\) 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*}

Solving the integral using 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 Figure 3.27 (with \(a=3\)), \(\theta = \arctan(x/3)\) and:

\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*}

🔗

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

🔗

Example 3.29.

To compute \(\ds \int \frac{dx}{x\sqrt{x^2-1}}\text{,}\) we consider the two intervals of the domain. For \(x \gt 1\text{,}\) let \(x = \sec(\theta)\) with \(0 \lt \theta \lt \pi/2\) (see 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. \quad (x \gt 1) \end{align*}

For \(x \lt -1\text{,}\) where \(\pi/2 \lt \theta \lt \pi\text{,}\) we have \(|\tan(\theta)| = -\tan(\theta)\text{,}\) yielding \(-\asec(x)+C\text{.}\) Combining these:

\begin{equation*} \int \frac{dx}{x\sqrt{x^2-1}} = g(x) +C \end{equation*}

where

\begin{align*} g(x) &= \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*}

The graph below shows \(g(x)\) and \(\arctan(\sqrt{x^2-1})\) differ by a constant on each connected component (\(0\) on \(x \ge 1\text{,}\) \(\pi\) on \(x \le -1\)).

🔗

See Exercise 3.5.7 for an alternative substitution.

🔗

The graph of g x together with the graph of arctan of square of x square minus 1. — Figure 3.30. \(g(x)\) and \(\arctan(\sqrt{x^2-1})\)
🔗

🔗

Next, an example with \(\sqrt{a^2-x^2}\text{.}\)

🔗

Example 3.31.

To compute \(\ds \int \frac{dx}{x\sqrt{1-x^2}}\text{,}\) let \(x = \sin(\theta)\) with \(-\pi/2 \le \theta \le \pi/2\) (see Figure 3.25). 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 gives \(4x^2 + 8x -5= 4(x+1)^2 - 9\text{.}\) Let \(u=2(x+1)\text{.}\) The integral becomes:

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

The integrand \(f(u) = 1/\sqrt{u^2-9}\) is even. If \(F(u)\) is an antiderivative for \(u \gt 3\text{,}\) then \(-F(-u)\) is one for \(u \lt 3\text{.}\) For \(u \gt 3\text{,}\) use \(u=3\sec(\theta)\text{:}\)

\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*}

Extending to the full domain by symmetry (as \(\frac{F(u)-F(-u)}{2}\) provides an odd antiderivative):

\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| = \ln|u+\sqrt{u^2-9}|-\ln(3). \end{align*}

Thus, the general solution is:

\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*}

🔗

Example 3.33.

To compute

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

let \(x =4\sin \theta\text{.}\) Then \(dx = 4\cos \theta d\theta\text{.}\) From the reference triangle:

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

Substituting:

\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*}

Using \(w = \tan \theta\text{:}\)

\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