∫(270-0)cos(2次)x/4d(x)

∫270cosxcosx/4dx
=270/4[sinxcosx+∫sinxsinxdx]
=270/4[sinxcox+1-∫cosxcosxdx]
∫270cosxcosx/4dx
=270/4[sinxcox+1]/2
=270/8[sinxcox+1]
=135/4[sinxcox+1]

∫(270-0)cos(2次)x/4d(x)
=∫(3π/2-0)cos(2次)x/4d(x)
=(1/2)∫(3π/2-0)(1+cosx/2)d(x)
=(1/4)∫(3π/2-0)(1+cosx/2)d(x/2)
=(1/4)(∫(3π/2-0)1d(x/2)+∫(3π/2-0)(cosx/2)d(x/2))
=(1/4)(3π/2+sin3π/4)
=(1/4)(3π/2+√2/2)=(3π+√2)/8

∫（270-0）(cosx)^2/4dx
=(135/4)∫(cos2x+1)dx
=(135/8)∫(cos2x+1)d(2x)
=(135/8)(sin2x+2x)+C