# 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}