Question

please help with integration problem!
integration of dx/ (x^3(squareroot of (x^2 -9))

Answer 1

let x = 3sec (theta) ---> x^2 = 9 sec ^2 ( theta) ; and dx = 3 sec(theta)tan(theta) d(theta)

Answer 2

x^3 = 27 sec(theta)

Answer 3

x^2 = 9 sec^2(theta)----> x^2 -9 = 9 sec^2 (theta) -9 = 9 [sec^2(theta) -1} = 9 tan^2 (theta)

Answer 4

we have x^3 = 27 sec^3(theta)
we have sqr (x^2 -9) = 3 tan (theta)
we have dx = 3 sec(theta)tan(theta) d(theta)(I dont write theta anymore, you must know it is)
replace all to the stuff: integral of 3sec. tan d(theta)/ 27sec^3* 3 tan
simplify, we have: integral of 1d(theta)/27 sec^2 = 1/27 integral of cos^2 . and cos^2 = 1-2cos /2. replace again, we have 1/27 integral of 1/2 - cos(theta). take integral. It is easy to go .
hope this help