integral of "arccos (3x)" ?

Question

if S [(arccos (x)] dx = x * arccos[x - sqrt(1-x^2)] + c

Then would intergral of "arccos (3x)" be:

3x * arccos [3x - sqrt (1-(3x)^2)] + c ????

Answer 1

Integrating by parts

∫arccosx dx = x arccosx - √(1 - x²) +c . You have the [ ...] in the wrong place.

∫arccos3x dx

u = arccos3x dv/dx = 1 so du/dx = 3 / √(1 - 9x²) and v = x

∫arccos3x dx

= xarccos3x - ∫3x/√(1 - 9x²) dx

= xarccos3x - (1/3)√(1 - 9x²) + c

Answer 2

Report me if you like for being a *****, I officially don't care but Nateena is a bloody liar. 4% of her answers are best answers. Never been given a best answer? Take a long walk off a short cliff you lying rat. But that was a fairly good joke. Didn't make me laugh but it was original.

Answer 3

let y = f(ax)

dy/dx = df/d(ax) d(ax)/dx = a df/d(ax)

so integrating y = f(ax)/a

so integral = (1/3)*f(3x) or (1/3)*(3x * arccos [3x - sqrt (1-(3x)^2)]) + C

and not what you have mentioned