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
$$ \displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2
\ \ \longrightarrow $$
$$ (2)^2 + (opposite)^2 = (x-2)^2
\ \ \longrightarrow \ \ \ opposite = \sqrt{(x-2)^2-4} = \sqrt{x^2-4x} \ \ \longrightarrow $$
$$ \tan \theta = \displaystyle{ opposite \over adjacent }= \displaystyle{ \sqrt{x^2-4x} \over 2 } . \Big) $$
$$ = \ln \Big|\frac{x-2}{2} + \frac{\sqrt{x^2-4x}}{2} \Big| + C $$
Click HERE to return to the list of problems.