Exercise XIX number 15 - Calculus Made Easy

Question

$$ \text{Use substitution}\quad\frac{1}{x}=\frac{b}{a}cosh\;u\quad\text{to show that}\quad\\ \int\;\frac{dx}{x\sqrt{a^2-b^2x^2}}=\frac{1}{a}\ln\frac{a-\sqrt{a^2-b^2x^2}}{x}\;+\;C.\\ \text{My way:}\\ \text{Let}\quad\frac{1}{x}=\frac{b}{a}cosh\;u\quad\\ x=\frac{a}{b}\frac{1}{cosh\;u}\\ dx=-\frac{a}{b}\frac{sinh\;u}{cosh^2u}du\\ x^2=\frac{a^2}{b^2}\frac{1}{cosh^2u}\\ u=cosh^-(\frac{a}{bx})\\ \int\frac{\frac{b\;cosh\;u}{a}\cdot-\frac{a\;sinh\;u}{b\;cosh^2u}}{\sqrt{a^2-b^2\cdot\frac{a^2}{b^2\;cosh^2u}}}du=\\ =\int\frac{-\frac{sinh\;u}{cosh\;u}=-tanh\;u}{\sqrt{a^2-\frac{a^2}{cosh^2u}}=\sqrt{a^2(1-\frac{1}{cosh^2u})}=\sqrt{a^2\;tanh^2u}=a\;tanh\;u}du=\\ =\int-\frac{du}{a}=\quad-\frac{1}{a}\int du=\quad -\frac{u}{a}\quad+\quad C =\quad-\frac{1}{a}cosh^-(\frac{a}{bx})\quad+\quad C=\\ =\quad-\frac{1}{a}\ln\Biggr(\frac{a}{bx}\pm\sqrt{\frac{a^2}{b^2x^2}-1}\Biggr)\quad+\quad C=\\ =\quad-\frac{1}{a}\ln\Biggr(\frac{a\pm\sqrt{a^2-b^2x^2}}{bx}\Biggr)\quad+\quad C=\\ =\quad-\frac{1}{a}\biggr(\ln(a\pm\sqrt{a^2-b^2x^2})-\ln(bx)\biggr)\quad+\quad C=\\ =\quad\frac{1}{a}\biggr(\ln(bx)-\ln(a\pm\sqrt{a^2-b^2x^2})\biggr)\quad+\quad C=\\ =\quad\frac{1}{a}\ln\frac{bx}{a\pm\sqrt{a^2-b^2x^2}}\quad+\quad C\\\text{What went wrong?} $$

Answer 1

You only have a minor error: $\cosh^-(\frac{a}{bx})= \ln\left(\frac{a}{bx} + \sqrt{\frac{a^2}{b^2x^2} - 1}\right)$, the minus-sign is not correct. This gives you $$\int\;\frac{dx}{x\sqrt{a^2-b^2x^2}}\quad = \quad\frac{1}{a}\ln\frac{bx}{a+\sqrt{a^2-b^2x^2}}+ C \\ = \quad \frac{1}{a}\ln\frac{b^2x}{a+\sqrt{a^2-b^2x^2}} - \frac{1}{a}\ln b + C $$ But $$\frac{b^2x}{a+\sqrt{a^2-b^2x^2}} \quad = \quad\frac{b^2x(a-\sqrt{a^2-b^2x^2})}{(a+\sqrt{a^2-b^2x^2})(a-\sqrt{a^2-b^2x^2})} \\ = \quad\frac{b^2x(a-\sqrt{a^2-b^2x^2})}{a^2 -(a^2-b^2x^2)} \quad = \quad\frac{a-\sqrt{a^2-b^2x^2}}{x}$$