Question

一道求不定积分的题目

∫cos^4xdx

Answer 1

cos^4x
=(cos²x)²
=[(1+cos2x)/2]²
=[1+2cos2x+(1+cos4x)/2]/4
=3/8+cos2x/2+cos4x/8
所以原式=∫3/8dx+1/2∫cos2xdx+1/8∫cos4xdx
=3x/8+1/4∫cos2xd2x+1/32∫cos4xd4x
=3x/8+1/4(sin2x)+1/32(sin4x)+C

Answer 2

cos^4x
=cos^2x*cos^2x
=cos^2x*(1-sin^2x)
=cos^2x-(sinxcosx)^2
=(1+cos2x)/2-(sin^2(2x))/4
=(1+cos2x)/2-(1-cos4x)/8
=(cos4x)/8+(cos2x)/2+3/8
所以
∫cos^4xdx
=∫((cos4x)/8+(cos2x)/2+3/8)dx
=(sin4x)/32+(sin2x)/4+3x/8+c