Michael Bowen's VC Course Pages

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Math V21B Section 7.5

Introduction

Please try this problem on your own before looking at my solution. There may be other ways to solve this; I make no claim that this method is the quickest, prettiest, or most efficient.

Problem 69

The original problem is $$I = \int_1^{\sqrt 3 } {\frac{{\sqrt {1 + {x^2}} }}{{{x^2}}}\;dx} $$

From the square root, we suspect a trigonometric substitution might be the best method. Using $$x = \tan \;\theta ;\quad dx = {\sec ^2}\theta \;d\theta ;\quad a^2 = 1$$ and making appropriate changes to the limits of integration, we can write $$I = \int_{{{\tan }^{ - 1}}1}^{{{\tan }^{ - 1}}\sqrt 3 } {\frac{{\sqrt {1 + {{\tan }^2}\theta } }}{{{{\tan }^2}\theta }}\;{{\sec }^2}\theta \;d\theta } = \int_{\pi /4}^{\pi /3} {\frac{{\sqrt {{{\sec }^2}\theta } }}{{{{\tan }^2}\theta }}\;{{\sec }^2}\theta \;d\theta }$$ $$I = \int_{\pi /4}^{\pi /3} {\frac{{{{\sec }^3}\theta }}{{{{\tan }^2}\theta }}d\theta }$$

Converting all of the tangents and secants to sines and cosines gives an expression in which all the trig functions are in the denominator, so this approach (which works with some integrals containing trig functions in a fraction) is unlikely to be productive immediately. Another possible approach is to use the identity ${\sec ^2}\theta = 1 + {\tan ^2}\theta$ to convert some secants to tangents, with hopes of producing an obvious $u$-substitution. This approach yields (after distributing the parentheses in the numerator of the first integral below) $$I = \int_{\pi /4}^{\pi /3} {\frac{{\left( {1 + {{\tan }^2}\theta } \right)\sec \theta }}{{{{\tan }^2}\theta }}d\theta } = \int_{\pi /4}^{\pi /3} {\frac{{\sec \theta }}{{{{\tan }^2}\theta }}d\theta } + \int_{\pi /4}^{\pi /3} {\frac{{{{\tan }^2}\theta \sec \theta }}{{{{\tan }^2}\theta }}d\theta }$$ Converting the secants and tangents in the first integral to sines and cosines is more productive now: $$I = \int_{\pi /4}^{\pi /3} {\frac{1}{{\cos \theta }}\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}d\theta } + \int_{\pi /4}^{\pi /3} {\sec \theta \;d\theta }$$ $$I = \int_{\pi /4}^{\pi /3} {\frac{{\cos \theta }}{{{{\sin }^2}\theta }}d\theta } + \int_{\pi /4}^{\pi /3} {\sec \theta \;d\theta }$$ Success! The first integral is a straightforward $u$-substitution, and the second integral is one we have seen often before, and whose result we have therefore recorded on our cheat sheet. The most effective $u$-substitution for the first integral is $$u = \sin \theta ;\quad du = \cos \theta \;d\theta$$ along with another change of limits, giving $$I = \int_{\sqrt 2 /2}^{\sqrt 3 /2} {\frac{{du}}{{{u^2}}}} + \int_{\pi /4}^{\pi /3} {\sec \theta \;d\theta } = \int_{\sqrt 2 /2}^{\sqrt 3 /2} {{u^{ - 2}}du} + \int_{\pi /4}^{\pi /3} {\sec \theta \;d\theta }$$ Evaluating the integrals, using the power rule for the first and the previously tabulated result for the second, yields $$I = \left. {\frac{{{u^{ - 1}}}}{{ - 1}}} \right|_{\sqrt 2 /2}^{\sqrt 3 /2} + \left. {\ln \left| {\sec \theta + \tan \theta } \right|} \right|_{\pi /4}^{\pi /3} = - \left. {\frac{1}{u}} \right|_{\sqrt 2 /2}^{\sqrt 3 /2} + \left. {\ln \left| {\sec \theta + \tan \theta } \right|} \right|_{\pi /4}^{\pi /3}$$

It's time for some arithmetic: $$I = - \left[ {\frac{1}{{\left( {\sqrt 3 /2} \right)}} - \frac{1}{{\left( {\sqrt 2 /2} \right)}}} \right] + \left[ {\ln \left| {\sec \frac{\pi }{3} + \tan \frac{\pi }{3}} \right| - \ln \left| {\sec \frac{\pi }{4} + \tan \frac{\pi }{4}} \right|} \right]$$ $$I = - \left[ {\frac{2}{{\sqrt 3 }} - \frac{2}{{\sqrt 2 }}} \right] + \left[ {\ln \left| {2 + \sqrt 3 } \right| - \ln \left| {\sqrt 2 + 1} \right|} \right] = \boxed{\sqrt 2 - \frac{2}{{\sqrt 3 }} + \ln \left( {2 + \sqrt 3 } \right) - \ln \left( {\sqrt 2 + 1} \right)}$$ The boxed answer at the end appears to agree with the solution in the back of the textbook.