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Re: Integral calculation!
Posted:
Mar 30, 2013 10:42 PM


On Sat, 30 Mar 2013, José Carlos Santos wrote: > On 30/03/2013 09:32, Ray Vickson wrote: > > > > I need to calculate integral(0,pi/4) x.tan x dx. > > > http://at.yorku.ca/cgibin/bbqa?forum=calculus;task=list > > > > Maple gets > > (1/8)*pi*ln(2) + (1/2)*Catalan, > > where Catalan is Catalan's constant, equal to > > Catalan = sum{(1)^n 1/(2n+1)^2,n=0..infinity} = 0.9159655942. > > > > Numerically, the integral is about 0.1857845357, > > Mathematica gets the same thing. Of course, this suggests that x.tan(x) > doesn't have an elementary primitive.
I agree and think it all starts with
integral(0,pi/2) x/sin x * dx = 2.sum(n=1,oo) 1/(2n1)^2 = k
From that I thought to get integral(0,pi/2) x/cos x * dx integral(0,pi/2) x/sin x * dx . . = integral(pi/2,0) (pi/2  x)/sin(pi/2  x) . . = integral(0,pi/2) (x  pi/2)/cos x . . = integral(0,pi/2) x/cos x  pi/2 * integral(0,pi/2) sec x dx . . = integral(0,pi/2) x/cos x  pi/2 * sec x tan x_0^pi/2 . . = integral(0,pi/2) x/cos x  pi/2 * sin x/cos^2 x_0^pi/2
But failed. If integral(0,pi/2) x/cos x * dx is known can one use integration by parts for integral(0,pi/2) x/cos x * sin x dx to make the conclusion?
 integral dfg = integral f.dg + integral g.df
lim(x>pi/2) x/cot x = lim(x>pi/2) 1/csc^2 x . . = lim(x>pi/2) sin^2 x = 1
x.tan x_0^pi/2 = pi/2




