2017-04-24 18:15:52 -0500 received badge ● Notable Question (source) 2017-03-19 04:34:10 -0500 received badge ● Notable Question (source) 2015-01-14 03:31:08 -0500 received badge ● Popular Question (source) 2014-09-08 10:16:03 -0500 received badge ● Popular Question (source) 2013-06-17 18:35:12 -0500 received badge ● Teacher (source) 2013-06-17 18:35:12 -0500 received badge ● Self-Learner (source) 2013-06-17 18:01:46 -0500 answered a question integrate cos(x)*cos(2x)*...*cos(mx) via SAGE So the answer is - Yes. It is a bug in algorithm='maxima', so use algorithm='mathematica_free' (def new function to find definite integral) or simplify_full() for such product of cos(kx) and than integrate. 2013-06-17 17:51:12 -0500 commented question integrate cos(x)*cos(2x)*...*cos(mx) via SAGE Thanks! Really if use simplify_full() than answer is correct 2013-06-17 17:49:45 -0500 commented question How to use algorithm='mathematica free' to calculate definite integral? Thanks for comments! 2013-06-17 17:49:19 -0500 answered a question How to use algorithm='mathematica free' to calculate definite integral? Thanks for comments! x=var('x') f = lambda x : x^2 def defintegral_viaMathFree(f,x,a,b): F(x)=integrate(f(x),x,algorithm='mathematica_free') return F(b)-F(a) And defintegral_viaMathFree(f,x,0,1) output: 1/3 2013-06-17 17:09:56 -0500 received badge ● Nice Question (source) 2013-06-17 14:57:15 -0500 received badge ● Student (source) 2013-06-17 08:46:01 -0500 asked a question How to use algorithm='mathematica free' to calculate definite integral? Is it possible to use algorithm='mathematica free' to calculate definite integral ? integrate(x^2,x,algorithm='mathematica_free') Output: 1/3*x^3 and integrate(x^2,x,0,1,algorithm='mathematica_free') Output: 1/3*x^3 but I want 1/3 2013-06-17 06:42:05 -0500 received badge ● Editor (source) 2013-06-17 06:39:21 -0500 asked a question integrate cos(x)*cos(2x)*...*cos(mx) via SAGE I'm going to find $I_m=\int_0^{2\pi} \prod_{k=1}^m cos(kx){}dx$, where $m=1,2,3\ldots$ Simple SAGE code: x=var('x') f = lambda m,x : prod([cos(k*x) for k in range(1,m+1)]) for m in range(1,15+1): print m, numerical_integral(f(m,x), 0, 2*pi),integrate(f(m,x),x,0,2*pi).n() Output: 1 -1.47676658757e-16 0.000000000000000 2 -5.27735962315e-16 0.000000000000000 3 1.57079632679 1.57079632679490 4 0.785398163397 0.785398163397448 5 -2.60536121164e-16 0.000000000000000 6 -1.81559273097e-16 0.000000000000000 7 0.392699081699 0.392699081698724 8 0.343611696486 0.147262155637022 9 -1.72448482421e-16 0.294524311274043 10 -1.8747663502e-16 0.196349540849362 11 0.214757310304 0.312932080728671 12 0.190213617698 0.177941771394734 13 -1.30355375996e-16 0.208621387152447 14 -1.25168280013e-16 0.0859029241215959 15 0.138441766107 0.134223318939994 As you can see numerical answer is right, but result of integrate(...) is right for $m=1,2,\ldots,7$ and then there is some bug. We can print indefinite integral: for m in range(7,11+1): print 'm=',m print 'Indef_I_m=',integrate(f(m,x),x) And Output: m = 7 Indef_I_m = 1/16*x + 1/16*sin(2*x) + 1/32*sin(4*x) + 7/384*sin(6*x) + 7/512*sin(8*x) + 3/320*sin(10*x) + 5/768*sin(12*x) + 5/896*sin(14*x) + 1/256*sin(16*x) + 1/384*sin(18*x) + 1/640*sin(20*x) + 1/704*sin(22*x) + 1/1536*sin(24*x) + 1/1664*sin(26*x) + 1/1792*sin(28*x) m = 8 Indef_I_m = 3/128*x + 5/256*sin(2*x) + 1/32*sin(3*x) + 5/512*sin(4*x) + 5/768*sin(6*x) + 1/256*sin(8*x) + 1/256*sin(10*x) + 1/256*sin(12*x) + 1/256*sin(14*x) + 1/256*sin(16*x) + 7/2304*sin(18*x) + 3/1280*sin(20*x) + 5/2816*sin(22*x) + 1/768*sin(24*x) + 3/3328*sin(26*x) + 1/1792*sin(28*x) + 1/1920*sin(30*x) + 1/4096*sin(32*x) + 1/4352*sin(34*x) + 1/4608*sin(36*x) + 3/32*sin(x) m = 9 Indef_I_m = 3/64*x + 3/128*sin(2*x) + 23/768*sin(3*x) + 3/256*sin(4*x) + 3/640*sin(5*x) + 1/128*sin(6*x) + 5/1792*sin(7*x) + 5/2304*sin(9*x) + 3/2816*sin(11*x) + 1/832*sin(13*x) + 1/1280*sin(15*x) + 3/4352*sin(17*x) + 5/4864*sin(19*x) + 1/1344*sin(21*x) + 3/2944*sin(23*x) + 7/6400*sin(25*x) + 1/1152*sin(27*x) + 3/3712*sin(29*x) + 5/7936*sin(31*x) + 1/2112*sin(33*x) + 3/8960*sin(35*x) + 1/4736*sin(37*x) + 1/4992*sin(39*x) + 1/10496*sin(41*x) + 1/11008*sin ... 2011-11-14 02:12:00 -0500 received badge ● Supporter (source)