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 $x$.

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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 $$ \sec \theta = \displaystyle{ hypotenuse \over adjacent }= \displaystyle{ \sqrt{x^2+1} \over 1 } = \sqrt{x^2+1} . \Big) $$ $$ = -\displaystyle{ \sqrt{x^2+1} \over x } + \ln \Big|\sqrt{x^2+1} + x \Big| + C $$

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