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 $$(adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(2)^2 + (opposite)^2 = (x)^2 \ \ \longrightarrow \ \ \ opposite = \sqrt{x^2-4} \ \ \longrightarrow$$ $$\tan \theta = \displaystyle{ opposite \over adjacent } = \displaystyle{ \sqrt{x^2-4} \over 2} . \Bigg)$$ $$= \displaystyle { 2 \Big(\frac{x}{2}\Big)\Big(\frac{\sqrt{x^{2}-4}}{2}\Big) - 2 \ln \Big|\frac{x}{2} + \frac{\sqrt{x^{2}-4}}{2} \Big| + C }$$ $$= \displaystyle { \frac{x \sqrt{x^{2}-4}}{2} - 2 \ln \Big|\frac{x}{2} + \frac{\sqrt{x^{2}-4}}{2} \Big| + C }$$