SOLUTION 9: $\ \$ To integrate $\displaystyle{ \int { \sqrt{x^2+25} } \ dx }$ use the trig substitution $$x = 5 \tan \theta$$ so that $$dx = 5 \sec^{2} \theta \ d \theta$$ Substitute into the original problem, replacing all forms of $x$, getting $$\displaystyle {\int \sqrt{25 \tan^{2} \theta + 25} \cdot 5 \sec^{2} \theta \ d \theta }$$ $$= \displaystyle { 25 \int \sqrt{\tan^{2} \theta + 1} \ \sec^{2} \theta \ d \theta }$$ $$= \displaystyle { 25 \int \sqrt{\sec^{2} \theta} \ \sec^{2} \theta \ d \theta }$$ $$= \displaystyle { 25 \int \sec \theta \ \sec^{2} \theta \ d \theta }$$ $$= \displaystyle { 25 \int \sec^{3} \theta \ d \theta }$$ 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 $$\displaystyle { \int \sqrt{x^{2}-4} \ dx = \displaystyle { 4 \int \tan \theta \ \sec \theta \tan \theta \ d \theta } = 4 \Big( \frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln \Big|\sec \theta + \tan \theta \Big| \Big) } + C$$ $$= \displaystyle { 2 \sec \theta \tan \theta - 2 \ln |\sec \theta + \tan \theta|} + C$$ \vskp \noindent (Recall from {\it PROBLEM 2} \ that $\displaystyle { \int \sec^{3} \theta \ d \theta = \frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln |\sec \theta + \tan \theta| + C }$) \vskp $= \displaystyle { 25 (\frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln |\sec \theta + \tan \theta| + C) }$ \vskp \noindent \Big(We need to write our final answer in terms of $x$.

Since $x = 5 \tan \theta$ it follows that \vskp $\tan \theta = \displaystyle{ x \over 5 } = \displaystyle{ opposite \over adjacent }$ , \vskp \noindent and from the Pythagorean Theorem that \vskp $\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$ \vskp $(5)^2 + (x)^2 = (hypotenuse)^2 \ \ \longrightarrow \ \ \ hypotenuse = \sqrt{25+x^2} \ \ \longrightarrow$ \vskp $\sec \theta = \displaystyle{ hypotenuse \over adjacent }= \displaystyle{ \sqrt{25+x^2} \over 5 } .$ \Big) \vskp $= \displaystyle { 25(\frac{1}{2}(\frac{\sqrt{x^{2}+25}}{5})(\frac{x}{5}) - \frac{1}{2}\ln |\frac{\sqrt{x^{2}+25}}{5} + \frac{x}{5}| + C) }$ \vskp $= \displaystyle { 25(\frac{x \sqrt{x^{2}+25}}{50} - \frac{1}{2}\ln |\frac{\sqrt{x^{2}+25}}{5} + \frac{x}{5}| + C) }$ . \vskp \vskp Since $x = \tan \theta$ it follows that $$\tan \theta = \displaystyle{ x \over 1 } = \displaystyle{ opposite \over adjacent }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(1)^2 + (x)^2 = (hypotenuse)^2 \ \ \longrightarrow \ \ \ hypotenuse = \sqrt{x^2+1} \ \ \longrightarrow$$ $$\csc \theta = \displaystyle{ hypotenuse \over opposite }= \displaystyle{ \sqrt{x^2+1} \over x }$$ and $$\sec \theta = \displaystyle{ hypotenuse \over adjacent }= \displaystyle{ \sqrt{x^2+1} \over 1 } = \sqrt{x^2+1} . \Big)$$ $$= -\displaystyle{ \sqrt{x^2+1} \over x } + \ln \Big|\sqrt{x^2+1} + x \Big| + C$$