8.4 | Trigonometric Substitution


For integrals involving Use the substitution Use the identity
\(\sqrt{a^2-x^2}, \; a>0 \) \(x = a \sin \theta, \; \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2} \) \( 1- \sin^2 \theta = \cos^2 \theta \)
\(\sqrt{a^2+x^2}, \; a>0 \) \(x = a \tan \theta, \; \frac{-\pi}{2} < \theta < \frac{\pi}{2} \) \( 1+\tan^2 \theta = \sec^2 \theta \)
\(\sqrt{x^2-a^2}, \; a>0 \) \(x = a \sec \theta, \; 0 \leq \theta < \frac{\pi}{2} \text{ or } \frac{\pi}{2} < \theta \leq \pi \) \( \sec^2 \theta – 1 = \tan^2 \theta \)

Trigonometric Substitution

For integrals involving Use the substitution Use the identity
\(\sqrt{a^2-x^2}, \; a>0 \) \(x = a \sin \theta, \; \frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2} \) \( 1- \sin^2 \theta = \cos^2 \theta \)
\(\sqrt{a^2+x^2}, \; a>0 \) \(x = a \tan \theta, \; \frac{-\pi}{2} < \theta < \frac{\pi}{2} \) \( 1+\tan^2 \theta = \sec^2 \theta \)
\(\sqrt{x^2-a^2}, \; a>0 \) \(x = a \sec \theta, \; 0 \leq \theta < \frac{\pi}{2} \text{ or } \frac{\pi}{2} < \theta \leq \pi \) \( \sec^2 \theta – 1 = \tan^2 \theta \)

Trigonometric Substitution

  • For integrals involving \(\sqrt{a^2-x^2}, \; a>0 \)
    • use the subsitution
      $$x = a \sin \theta, $$
      where \(\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}\)
    • and the identity
      $$ 1- \sin^2 \theta = \cos^2 \theta $$
  • For integrals involving \(\sqrt{a^2+x^2}, \; a>0 \)
    • use the substitution
      $$x = a \tan \theta,$$
      where \(\frac{-\pi}{2} < \theta < \frac{\pi}{2}\)
    • and the identity
      $$ 1+\tan^2 \theta = \sec^2 \theta $$
  • For integrals involving \(\sqrt{x^2-a^2}, \; a>0 \)
    • use the substitution
      $$x = a \sec \theta,$$
      where \(0 \leq \theta < \frac{\pi}{2}\) or \(\frac{\pi}{2} < \theta \leq \pi \)
    • and the identity
      $$ \sec^2 \theta – 1 = \tan^2 \theta $$
Find
$$
\int \frac{1}{(4+x^2)^2} \; dx
$$


Let \(x = 2\tan \theta \) then \( dx = 2\sec^2 \theta \; d \theta \) and

$$
\begin{aligned}
\int \frac{1}{(4+x^2)^2} \; dx & = \int \frac{1}{((4+ (2 \tan \theta )^2)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16(1+\tan^2 \theta)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16 \sec^4 \theta} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{8 \sec^2 \theta} \; d \theta = \frac{1}{8}\int \cos^2 \theta \; d \theta \\
& = \frac{1}{8}\int \frac{1+ \cos(2 \theta)}{2} \; d \theta \\
& = \frac{1}{16}\int 1+ \cos(2 \theta) \; d \theta \\
& = \frac{1}{16}\left( \theta+ \frac{\sin(2 \theta)}{2} \right) + C \\
& = \frac{1}{16}\left( \theta+ \frac{2\sin(\theta) \cos(\theta) }{2} \right) + C \\
\end{aligned}
$$
$$
\begin{aligned}
\int \frac{1}{(4+x^2)^2} \; dx & = \int \frac{1}{((4+ (2 \tan \theta )^2)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16(1+\tan^2 \theta)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16 \sec^4 \theta} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{8 \sec^2 \theta} \; d \theta = \frac{1}{8}\int \cos^2 \theta \; d \theta \\
& = \frac{1}{8}\int \frac{1+ \cos(2 \theta)}{2} \; d \theta \\
& = \frac{1}{16}\int 1+ \cos(2 \theta) \; d \theta \\
& = \frac{1}{16}\left( \theta+ \frac{\sin(2 \theta)}{2} \right) + C \\
& = \frac{1}{16}\left( \theta+ \frac{2\sin(\theta) \cos(\theta) }{2} \right) + C \\
\end{aligned}
$$
$$
\begin{aligned}
\int & \frac{1}{(4+x^2)^2} \; dx \\
& = \int \frac{1}{((4+ (2 \tan \theta )^2)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16(1+\tan^2 \theta)^2} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{16 \sec^4 \theta} 2\sec^2 \theta \; d \theta \\
& = \int \frac{1}{8 \sec^2 \theta} \; d \theta = \frac{1}{8}\int \cos^2 \theta \; d \theta \\
& = \frac{1}{8}\int \frac{1+ \cos(2 \theta)}{2} \; d \theta \\
& = \frac{1}{16}\int 1+ \cos(2 \theta) \; d \theta \\
& = \frac{1}{16}\left( \theta+ \frac{\sin(2 \theta)}{2} \right) + C \\
& = \frac{1}{16}\left( \theta+ \frac{2\sin(\theta) \cos(\theta) }{2} \right) + C \\
\end{aligned}
$$

Since \(x = 2 \tan \theta\), \( \tan \theta = \frac{x}{2} \) and \( \theta = \tan^{-1}\left( \frac{x}{2} \right) \). Also
$$
\sin \theta = \frac{x}{\sqrt{4+x^2}} \text{ and } \cos \theta = \frac{2}{\sqrt{4+x^2}}.
$$
So,

$$
\begin{aligned}
\int \frac{1}{(4+x^2)^2} \; dx & = \frac{1}{16}\left( \tan^{-1}\left( \frac{x}{2} \right)+ \frac{2\frac{x}{\sqrt{4+x^2}} \frac{2}{\sqrt{4+x^2}}}{2} \right) + C \\
& = \frac{1}{16} \left(\tan^{-1}\left( \frac{x}{2} \right)+ \frac{2x}{\sqrt{4+x^2}} \right) + C
\end{aligned}
$$
$$
\begin{aligned}
\int & \frac{1}{(4+x^2)^2} \; dx \\
& = \frac{1}{16}\left( \tan^{-1}\left( \frac{x}{2} \right)+ \frac{2\frac{x}{\sqrt{4+x^2}} \frac{2}{\sqrt{4+x^2}}}{2} \right) + C \\
& = \frac{1}{16} \left(\tan^{-1}\left( \frac{x}{2} \right)+ \frac{2x}{\sqrt{4+x^2}} \right) + C
\end{aligned}
$$
$$
\begin{aligned}
& \int \frac{1}{(4+x^2)^2} \; dx \\
& = \frac{1}{16}\left( \tan^{-1}\left( \frac{x}{2} \right)+ \frac{2\frac{x}{\sqrt{4+x^2}} \frac{2}{\sqrt{4+x^2}}}{2} \right) + C \\
& = \frac{1}{16} \left(\tan^{-1}\left( \frac{x}{2} \right)+ \frac{2x}{\sqrt{4+x^2}} \right) + C
\end{aligned}
$$

E 8.4 Exercises