∫（cos^4x-cos^6x ）dx=？

∫（cos^4x-cos^6x ）dx=？
∫（cos^4x-cos^6x ）dx
=∫cos^4x dx-∫cos^6xdx
(1)∫cos^4x dx
=1/4∫(1+cos2x)^2dx
=1/4∫(cos^2 2x+2cos2x+1)dx
=1/4(∫cos^2 2xdx+sin2x+x)
=1/4[1/2∫(1+cos4x)dx+sin2x+x]
=1/32sin4x+1/4sin2x+3/8x+C1
(2)因为cos²x=(1+cos2x)/2
所以：
cos^6x=[(1+cos2x)/2]³
=(1+3cos2x+3cos²2x+cos³2x)/8
cos²2x=(1+cos4x)/2
所以∫cos^6xdx
=1/8[∫(1+3cos2x)dx+3/2∫(1+cos4x)dx+∫cos³2xdx]
=1/8[(x+1.5sin2x)+1.5(x+(1/4)sin4x)+0.5∫(1-sin²2x)d（sin2x）]
=1/32(10x+6sin2x+1.5sin4x+2sin2x-(2/3)sin³2x)+C2
=1/96(30x+24sin2x+4.5sin4x-2sin³2x)+C2
=1/192(60x+48sin2x+9sin4x-4sin³2x)+C2
因此：
∫（cos^4x-cos^6x ）dx
=∫cos^4x dx-∫cos^6xdx
=1/32sin4x+1/4sin2x+3/8x+C1-[1/192(60x+48sin2x+9sin4x-4sin³2x)+C2]
=1/32sin4x+1/4sin2x+3/8x+C1-1/192(60x+48sin2x+9sin4x-4sin³2x)-C2
=1/32sin4x+1/4sin2x+3/8x-1/192(60x+48sin2x+9sin4x-4sin³2x)+C
=12/192x-3/192sin4x-1/48sin³2x+C
=(1/16)x-(1/64)sin4x-1/48sin³2x+C