Problem with the indefinite integral $\int{\frac{\sqrt{{{b}^{2}}{{x}^{2}}-{{a}^{2}}}}{{{x}^{2}}}dx}$

Question

I tried using a substitution $x=\frac{a}{b}\sec \theta $ to solve

$$\int{\frac{\sqrt{{{b}^{2}}{{x}^{2}}-{{a}^{2}}}}{{{x}^{2}}}dx}$$

Obviously, $dx=\frac{a}{b}\sec \theta \tan \theta d\theta $. In other words,

$\begin{align} \int{\frac{\sqrt{{{b}^{2}}{{x}^{2}}-{{a}^{2}}}}{{{x}^{2}}}dx} &=\int{\frac{\sqrt{{{b}^{2}}{{\left( \frac{a}{b}\sec \theta \right)}^{2}}-{{a}^{2}}}}{{{\left( \frac{a}{b}\sec \theta \right)}^{2}}}\left( \frac{a}{b}\sec \theta \tan \theta \right)d\theta } \\ &=\int{\frac{\sqrt{{{\sec }^{2}}\theta -1}}{\frac{1}{b}\sec \theta }\left( \tan \theta \right)d\theta } \\ & =b\int{\frac{{{\tan }^{2}}\theta }{\sec \theta }d\theta } \\ &=b\int{\frac{{{\sec }^{2}}\theta -1}{\sec \theta }d\theta } \\ \end{align}$

But I feel there’s something wrong with my solution. Perhaps there’s a more effective/better way to solve it?

Answer 1

HINT

You may try the substitution \begin{align*} x = \frac{a\cosh(z)}{b} \end{align*}

whence we get that \begin{align*} \int\frac{\sqrt{b^{2}x^{2} - a^{2}}}{x^{2}}\mathrm{d}x & = b\int\frac{\sqrt{\cosh^{2}(z) - 1}\sinh(z)}{\cosh^{2}(z)}\mathrm{d}z\\\\ & = b\int\frac{\sinh^{2}(z)}{\cosh^{2}(z)}\mathrm{d}z\\\\ & = b\int\frac{\cosh^{2}(z) - 1}{\cosh^{2}(z)}\mathrm{d}z\\\\ & = b\left[\int\mathrm{d}z - \int\frac{\mathrm{d}z}{\cosh^{2}(z)}\right]\\\\ & = b[z - \tanh(z)] + c\\\\ & = b\left[\cosh^{-1}\left(\frac{bx}{a}\right) - \frac{\sinh\left(\cosh^{-1}\left(\frac{bx}{a}\right)\right)}{\cosh\left(\cosh^{-1}\left(\frac{bx}{a}\right)\right)}\right] + c \end{align*} where you can use the relations: \begin{align*} \begin{cases} \cosh^{-1}(z) = \ln(z + \sqrt{z^{2} - 1})\\\\ \cosh^{2}(z) - \sinh^{2}(z) = 1\\\\ \cosh(\cosh^{-1}(z)) = z \end{cases} \end{align*}

Can you take it from here?

Answer 2

Hint

Use one integration by parts $$\int{\frac{\sqrt{{{b}^{2}}{{x}^{2}}-{{a}^{2}}}}{{{x}^{2}}}dx}=-\frac{\sqrt{b^2 x^2-a^2}}{x}+\int \frac{b^2}{\sqrt{b^2 x^2-a^2}}\,dx$$