cosx的4次方的不定积分

网友回答

原式=(1/4)∫（1+cos2x)^2dx
=(1/4)∫[1+2cos2x+(cos2x)^2]dx
=x/4+(sinx)/4+(1/8)∫(1+cos4x)dx
=x/4+(sinx)/4+x/8+(sin4x)/32+C
=3x/8+(sinx)/4+(sin4x)/32+C
======以下答案可供参考======
供参考答案1:
∫cos^4xdx=∫[(1+cos2x)/2]^2dx
=1/4∫[1+2cos2x+cos^2(2x)]dx
=1/4∫[1+2cos2x+(1+cos4x)/2]dx
=1/8∫(3+4cos2x+cos4x)dx
=1/8(3x+2sin2x+1/4*sin4x)+C