# Using the method of trigonometric substitution show that the definite integral `int_0^5 sqrt(x^2-16)/x^2 dx approx 0.09`

mathsworkmusic | Certified Educator

calendarEducator since 2012

starTop subjects are Math, Science, and Business

We want `int_0^5 sqrt(x^2-16)/x^2 dx = int_4^5 sqrt(x^2 -16)/x^2 dx` because the function is undefined below `x=4`

Make the substitution `x = 4secu` ` `

Then `x^2 -16 = 4^2(sec^2u -1) =...

(The entire section contains 92 words.)