Question 1027913
{{{int((x/(1+cos^2(x))), dx, 0, pi) = 2int((x/(3+cos2x)), dx, 0, pi) }}}

by using the identity {{{(1+cos2x)/2 = cos^2(x)}}}.

Now let {{{x = pi - y}}} and substitute into the second integral.

==> {{{2int((x/(3+cos2x)), dx, 0, pi)  = 2int(((pi-y)/(3+cos2y)), (-dy), pi, 0) = 2int(((pi-y)/(3+cos2y)), dy, 0, pi) = 2int((pi/(3+cos2y)), dy, 0, pi)-2int((y/(3+cos2y)), dy, 0, pi)}}}

={{{2*pi*int((1/(3+cos2x)), dx, 0, pi)-2int((x/(3+cos2x)), dx, 0, pi)}}}

==> {{{4int((x/(3+cos2x)), dx, 0, pi) = 2*pi*int((1/(3+cos2x)), dx, 0, pi)}}}

==>{{{int((x/(3+cos2x)), dx, 0, pi) = (pi/2)int((1/(3+cos2x)), dx, 0, pi)}}}

Now {{{int((1/(3+cos2x)), dx, 0, pi) = (1/2)int((1/(sin^2(x) + 2cos^2(x))), dx, 0,pi)}}} , by letting {{{3 =  3sin^2(x) + 3cos^2(x)}}} and {{{cos2x = cos^2(x) - sin^2(x)}}}.

==> {{{int( (1/(sin^2(x)+2cos^2(x))), dx,0,pi) = int( (csc^2(x)/(1+2cot^2(x))), dx,0,pi)  = (-1/sqrt(2))int((1/(1+(sqrt(2)*cotx)^2)), d(sqrt(2)*cotx), 0, pi)}}}

= {{{(-1/sqrt(2))(tan^-1(sqrt(2)*cotx))[0]^pi = (-1/sqrt(2))(tan^-1(-infinity) - tan^-1(infinity)) = (-1/sqrt(2))(-pi/2-pi/2) = pi/sqrt(2)}}}

==> {{{int((x/(3+cos2x)), dx, 0, pi) = (pi/4)*int( (1/(sin^2(x)+2cos^2(x))), dx,0,pi) = (pi/4)*(pi/sqrt(2)) = pi^2/(4sqrt(2))}}}


==> {{{int((x/(1+cos^2(x))), dx, 0, pi) = 2int((x/(3+cos2x)), dx, 0, pi) =2pi^2/(4sqrt(2)) = pi^2/(2sqrt(2)) }}}