SOLUTION 12: $\ \$ Integrate $\displaystyle{ \int { 1 \over \sqrt{x^2-4x} } \ dx }$. First complete the square. Then $$\displaystyle { \int \frac{1}{\sqrt{x^{2}-4x}} \ dx = \int \frac{1}{\sqrt{(x^{2}-4x + 4) - 4}} \ dx = \int \frac{1}{\sqrt{(x-2)^{2}-4}} \ dx}$$ Use the trig substitution $$x-2 = 2 \sec \theta$$ so that $$dx = 2 \sec \theta \tan \theta \ d \theta$$ Substitute into the original problem, replacing all forms of x, getting $$\displaystyle { \int \frac{1}{\sqrt{(x-2)^{2}-4}} \ dx } = \displaystyle { \int \frac{1}{\sqrt{4 \sec^{2} \theta - 4}} \cdot 2 \sec \theta \tan \theta \ d \theta}$$ $$= \displaystyle { \int \frac{1}{\sqrt{4 (\sec^{2} \theta - 1)}} \cdot 2 \sec \theta \tan \theta \ d \theta}$$ $$= \displaystyle { \int \frac{1}{\sqrt{4 \tan^{2} \theta}} \cdot 2 \sec \theta \tan \theta \ d \theta}$$ $$= \displaystyle { \int \frac{1}{2 \tan \theta} \cdot 2 \sec \theta \tan \theta \ d \theta}$$ $$= \displaystyle { \int \sec \theta \ d \theta }$$ $$= \ln \Big|\sec \theta + \tan \theta \Big| + C$$ $\Big($We need to write our final answer in terms of $x$.

Since $x-2 = 2 \sec \theta$ it follows that $$\sec \theta = \displaystyle{ x-2 \over 2 } = \displaystyle{ hypotenuse \over adjacent }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(2)^2 + (opposite)^2 = (x-2)^2 \ \ \longrightarrow \ \ \ opposite = \sqrt{(x-2)^2-4} = \sqrt{x^2-4x} \ \ \longrightarrow$$ $$\tan \theta = \displaystyle{ opposite \over adjacent }= \displaystyle{ \sqrt{x^2-4x} \over 2 } . \Big)$$ $$= \ln \Big|\frac{x-2}{2} + \frac{\sqrt{x^2-4x}}{2} \Big| + C$$