Answer to Question #103227 in Calculus for Patrick Kariuki

Question

Integrate dx/(25x²-4)½ using trigonometric substitution

Answer 1

Substitution:

$x=\frac{2}{5}\sec{\theta}$

Then $dx=\frac{2}{5}\frac{\sin{\theta}}{\cos^2{\theta}}d\theta$

$\int\frac{dx}{\sqrt{25x^2-4}}=\int\frac{2}{5}\frac{\sin{\theta}}{\cos^2{\theta}\sqrt{4\sec^2{\theta}-4}}d\theta=\frac{1}{5}\int\frac{\sin{\theta}}{\cos^2{\theta}\sqrt{\sec^2{\theta}-1}}d\theta=$

$=\frac{1}{5}\int\frac{\sin{\theta}}{\cos^2{\theta}\tan{\theta}}d\theta=\frac{1}{5}\int\frac{d\theta}{\cos{\theta}}=\frac{1}{5}\int\sec{\theta}d\theta=\frac{1}{5}\ln|\sec{\theta}+\tan{\theta}|+C.\\ http://www.math.com/tables/integrals/more/sec.htm$

$\sec{\theta}=\frac{5}{2}x$

$x^2=\frac{4}{25}\frac{1}{\cos^2{\theta}}\\ x^2-\frac{4}{25}=\frac{4}{25}tan^2{\theta}\\ tan^2{\theta}=\frac{25}{4}x^2-1\\ tan{\theta}=\sqrt{\frac{25}{4}x^2-1}$

We have

$\int\frac{dx}{\sqrt{25x^2-4}}=\frac{1}{5}\ln|\frac{5}{2}x+\sqrt{\frac{25}{4}x^2-1}|+C.$