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اوجد تكامل جا^4(س)

السبت، 18 فبراير 2012 التسميات:
من المتطابقة المشهورة : جا²س = ½(1 - جتا(2س)) ، جتا²(س) = ½(1 + جتا(2س)
نستطيع الحصول على :
∫جا^4(س) دس

= ∫[½(1 - جتا(2س) ]² دس

= 1\4∫[(1 - 2جتا(2س) + جتا²(2س) ] دس

= ∫(1\4 - ½جتا(2س) + 1\4(½(1+جتا(4س)) دس

= 1\4∫دس - ½∫جتا(2س) دس + 
1/8 ∫دس + 1/8 ∫جتا(4س) دس

= 1\4س - 1\4جا(2س) + 1\8س + 1\32جا(4س) + ث

= 3\8س - 1\4جا(2س) + 1\32جا(4س) + ث

►الحل بالرموز الإنجليزية◄ 

= 1/4∫(1 - 2cos(2x) + cos²(x) ) dx

= 1/4∫dx - ½∫cos(2x)dx + 1/4∫cos²(x)dx

1/4X - 1/4sin(2X) + 1/8∫(1+cos(4x)

=1/4X - 1/4sin(2X) + 1/8∫dx + 1/8∫cos(4x)dx

=1/4X - 1/4sin(2X) + 1/8 X + 1/32 sin(4X) + C
 
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