SOLUTION 10: $ \ \ $ To integrate $ \displaystyle{ \int {\sqrt{x^2-4} } \ \ dx } $ use the trig substitution
$$ x = 2 \sec \theta $$
so that
$$ dx = 2 \sec \theta \tan \theta \ \ d\theta $$
Substitute into the original problem, replacing all
forms of $ x $, getting
$$ \displaystyle { \int \sqrt{x^{2}-4} \ dx} = \displaystyle { \int
\sqrt{ 4 \sec^{2} \theta - 4 } \cdot 2 \sec \theta \tan \theta \ d
\theta } $$
$$ = \displaystyle{ 2 \int \sqrt{4(\sec^{2} \theta - 1) } \ \sec
\theta \tan \theta \ d \theta } $$
$$ = \displaystyle{ 4 \int \sqrt{\sec^{2} \theta - 1 } \ \sec
\theta \tan \theta \ d \theta } $$
$$ = \displaystyle { 4 \int \sqrt{\tan^{2} \theta} \ \sec \theta
\tan \theta \ d \theta } $$
$$ = \displaystyle { 4 \int \tan \theta \ \sec \theta \tan \theta
\ d \theta } $$
Now use integration by parts. Let $ u = \tan \theta $ and $ dv = \sec \theta \tan \theta \ d \theta $ , so that $ du = \sec^2 \theta \ d \theta $ and $ v = \sec \theta $. Let $ A = \displaystyle { \int \tan \theta \sec \theta \tan \theta \ d \theta } $. Then
$$ A = \displaystyle { \int \tan \theta \sec \theta \tan \theta \ d \theta } = \sec \theta \tan \theta - \displaystyle { \int \sec \theta \sec^{2} \theta \ d \theta } $$
(Recall that $ \sec^2 \theta = 1 + \tan^2 \theta $.)
$$ = \displaystyle { \sec \theta \tan \theta - \int \sec \theta
(1 + \tan^{2} \theta) \ d \theta } $$
$$ = \displaystyle { \sec \theta \tan \theta -
\int \sec \theta \ d \theta - \int \sec \theta \tan^{2} \theta \ d \theta } $$
(Recall that $ \displaystyle { \int \sec \theta
\ d \theta } = \ln\Big| \sec \theta + \tan \theta \Big| + C $.)
$$ = \displaystyle { \sec \theta \tan \theta -
\ln\Big| \sec \theta + \tan \theta \Big| - \int \tan \theta \sec \theta \tan \theta) \ d \theta } $$
$$ = \displaystyle { \sec \theta \tan \theta -
\ln\Big| \sec \theta + \tan \theta \Big| - A } $$
Then
$$ 2A = \sec \theta \tan \theta - \ln\Big| \sec \theta + \tan \theta \Big| + C $$
so that
$$ \displaystyle { A = \int \tan \theta \sec \theta \tan \theta \ d \theta = \frac{1}{2}
\sec \theta \tan \theta - \frac{1}{2} \ln \Big|\sec \theta + \tan
\theta \Big| + C } $$
and
$$ \displaystyle { \int \sqrt{x^{2}-4} \ dx
= \displaystyle { 4 \int \tan \theta \ \sec \theta \tan \theta
\ d \theta }
= 4 \Big( \frac{1}{2} \sec \theta
\tan \theta - \frac{1}{2} \ln \Big|\sec \theta + \tan \theta \Big| \Big) } + C $$
$$ = \displaystyle { 2 \sec \theta \tan \theta - 2 \ln \Big|\sec \theta
+ \tan \theta \Big|} + C $$
$ \Bigg( $ We need to write our final answer in terms of $x$.
Since
$$ x = 2 \sec \theta $$
it follows that
$$ \sec \theta = \displaystyle{ x \over 2 } = \displaystyle{ hypotenuse \over adjacent } $$
and from the Pythagorean Theorem that
$$ (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow $$
$$ (2)^2 + (opposite)^2 = (x)^2 \ \ \longrightarrow \ \ \ opposite = \sqrt{x^2-4} \ \ \longrightarrow $$
$$ \tan \theta = \displaystyle{ opposite \over adjacent } = \displaystyle{ \sqrt{x^2-4} \over 2} . \Bigg) $$
$$ = \displaystyle { 2 \Big(\frac{x}{2}\Big)\Big(\frac{\sqrt{x^{2}-4}}{2}\Big) - 2
\ln \Big|\frac{x}{2} + \frac{\sqrt{x^{2}-4}}{2} \Big| + C } $$
$$ = \displaystyle { \frac{x \sqrt{x^{2}-4}}{2} - 2 \ln
\Big|\frac{x}{2} + \frac{\sqrt{x^{2}-4}}{2} \Big| + C } $$
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