Michael Bowen's VC Course Pages
Answers
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 29
The original problem is $$I = \int {\ln \left( {x + \sqrt {{x^2} - 1} } \right)\;dx} $$
From the square root, we suspect a trigonometric substitution might be the best method; if we set $a^2=1$, the radicand (expression inside the square root) can be seen to be of the form $\sqrt {{x^2} - {a^2}}$. This suggests the substitution $x = a\sec \theta$ or, since $a=1$, $$x = \sec \theta$$ $$dx = \sec \theta \tan \theta \;d\theta$$ Rewriting the original problem makes it more clear how to place the above substitutions: $$I = \int {\ln \left( {\sec \theta + \sqrt {{{\sec }^2}\theta - 1} } \right)\sec \theta \tan \theta \;d\theta } = \int {\ln \left( {\sec \theta + \sqrt {{{\tan }^2}\theta } } \right)\sec \theta \tan \theta \;d\theta }$$ $$I = \int {\ln \left( {\sec \theta + \tan \theta } \right)\sec \theta \tan \theta \;d\theta } $$
The logarithm expression should look familiar; it is the antiderivative of the secant function. This would suggest a $u$-substitution, with $$u = \ln \left( {\sec \theta + \tan \theta } \right)$$ $$du = \sec \theta \;d\theta$$ which would be perfect except for the pesky extra factor of $\tan \theta$ in the integrand; alas, this one extra factor forces us to discard the $u$-substitution approach. With no other obvious $u$-substitutions available, we note that $\ln \left( {\sec \theta + \tan \theta } \right)$ is the antiderivative of something simple, while ${\sec \theta \tan \theta }$ is the derivative of something simple. Such a combination of relationships suggests the possibility of integration by parts, with $u$ and $du$ set as suggested above; it would follow that $$dv = \sec \theta \tan \theta \;d\theta $$ $$v = \sec \theta $$ With the integration-by-parts variables set up in this way, we may write $$I = \int {u\;dv} = uv - \int {v\;du} = \sec \theta \cdot \ln \left( {\sec \theta + \tan \theta } \right) - \int {\sec \theta \cdot \sec \theta \;d\theta } $$ $$I = \sec \theta \cdot \ln \left( {\sec \theta + \tan \theta } \right) - \int {{{\sec }^2}\theta \;d\theta } = \sec \theta \cdot \ln \left( {\sec \theta + \tan \theta } \right) - \tan \theta + C$$
From here, it is necessary to back-substitute for $x$ (using a right triangle where $x=\sec \theta$, or $\frac{1}{x}=\cos \theta$). Finding the missing leg of this triangle (using the Pythagorean theorem) leads us to conclude that $\tan \theta = \sqrt {{x^2} - 1}$. The final result, written in terms of $x$, is therefore $$\boxed{\displaystyle{I = x\ln \left( {x + \sqrt {{x^2} - 1} } \right) - \sqrt {{x^2} - 1} + C}}$$
