Use the trigonometric substitution to evaluate integrals involving the radicals,

$$ \sqrt{a^2 - x^2} , \ \ \sqrt{a^2 + x^2} , \ \ \sqrt{x^2 - a^2} $$

Case I: $\sqrt{a^2 - x^2}$

substitution: $\color{red}{x = a \sin \theta}$ $$ \sqrt{a^2 - x^2} = \sqrt{a^2 - (a \sin \theta)^2} = a \sqrt{1 - (\sin \theta)^2} = a \cos \theta $$

Example 1:

$$ \int \frac{dx}{x^2 \sqrt{9 - x^2}} $$

Solution:

substitution: $\color{red}{x = 3 \sin \theta}$ $$ \sqrt{9 - x^2} = 3 \cos \theta $$

Then,

$$ \begin{aligned} \int \frac{dx}{x^2 \sqrt {9 - x^2}} &= \int \frac{d (3 \sin \theta)}{[ 3 \sin \theta]^2 \cdot 3 \cos \theta} = \int \frac{3 \cos \theta d \theta}{27 \sin^2 \theta \cos \theta} = \\ &= \frac{1}{3} \int \frac{1}{sin^2 \theta} d \theta = \frac{- \cot \theta}{9} + C = \\ &= \frac{-\sqrt{9 - x^2}}{9x} + C \end{aligned} $$

Example 2:

$$ \int \frac{dx}{x^2 \sqrt{4 - x^2}} $$

Solution:

Let $x = 2 \sin \theta$, $dx = 2 \cos \theta d \theta$

$$ \int \frac{2 \cos \theta d \theta}{4 \sin^2 \theta \sqrt{4 - 4 \sin^2 \theta}} = \int \frac{d \theta}{4 \sin^2 \theta} = \frac{1}{4} \int \csc^2 \theta \ \ \ d \theta = - \frac{1}{4} \cot \theta + C $$

Case II: $\sqrt{a^2 + x^2}$

substitution: $\color{red}{x = a \tan \theta}$

Example 3:

$$ \int \frac{dx}{(\sqrt{x^2 + 1})^3} $$

Solution:

substitution: $x = \tan \theta$ $$ \sqrt{1 + x^2} = \sec \theta $$

Then,

\begin{aligned} \int \frac{dx}{(\sqrt{x^2 + 1})^3} &= \int \frac{d \tan \theta}{\sec^3 \theta} = \int \frac{\sec^2 \theta}{sec^3 theta} dx = \\ &= \int \frac{1}{\sec \theta} d \theta = \int \cos \theta d \theta = \sin \theta + C = \\ &= \frac{x}{\sqrt{1 + x^2}} + C \end{aligned}

Case III: $\sqrt{x^2 - a^2}$

substitution: $\color{red}{x = a \sec \theta}$ $$ \sqrt{a^2 + x^2} = \sqrt{a^2 + (a \tan \theta)^2} = a \sqrt{1 + (\sin \theta)^2} = a \sec \theta $$

Example 4:

$$ \int \frac{dx}{(\sqrt{x^2 - 4x})^3} $$

Solution:

$$ x^2 - 4x = x^2 - 4x + 4 - 4 = (x - 2)^2 - 2^2 \\ \leftrightarrow \frac{x - 2}{2} = \sec \theta \leftrightarrow x = 2 + 2 \sec \theta \\ \leftrightarrow \frac{\sqrt{x^2 - 4x}}{2} = \tan \theta \leftrightarrow \sqrt{x^2 - 4} = 2 \tan \theta $$

Then,

_{$$ \begin{aligned} \int \frac{dx}{(\sqrt{x^2 - 4x})^2} &= \int \frac{d (2 + 2 \sec \theta)}{[2 \tan \theta]^3} = \\ &= \int \frac{2 \sec \theta \tan \theta}{8 \tan^3 \theta}d \theta = \frac{1}{4} \int \frac{\sec \theta}{\sin^2 \theta} d \theta = \\ &= \frac{1}{4} \int \frac{1}{\cos \theta} \cdot \frac{\cos^2 \theta}{\sin^2 \theta} d \theta = \frac{1}{4} \int \frac{\cos \theta}{\sin^2 \theta} d \theta = \\ &= \frac{1}{4} \int \frac{1}{\sin^2 \theta} d \sin \theta \mathop = \limits^{u = \sin \Theta } \frac{1}{4} \int \frac{1}{u^2} du = \\ &= \frac{-1}{4u} + C = \frac{-1}{4 \sin \theta} + C = \frac{-1}{4 \cdot \left( \frac{\sqrt{x^2 - 4x}}{x-2} \right)} + C = \\ &= \frac{- (x - 2)}{4 \sqrt{x^2 - 4x}} + C \end{aligned} $$}