The discussion focuses on evaluating the integral of cos-1(2x) dx using integration by parts. The correct approach involves setting u = cos-1(2x) and dv = dx, leading to the derivative du = -2/(sqrt(1 - (2x)2)) dx. The integration by parts formula is applied, resulting in the expression u * v - ∫v du, where v is the integral of dx. The participants clarify that the initial misunderstanding involved incorrectly multiplying arccos by 2x.
PREREQUISITESStudents studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of integration by parts involving inverse functions.
crazco said:The Attempt at a Solution
let
u = arc cos
du = -1 / (sqrt 1-x^2)
dv = 2x
v= x^2
arc cos * x^2 - the integral of -x^2 / (sqrt 1-x^2)
then i don't know what to do