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 $x$.

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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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