Question

SIN^4XCOS^4XD的不定积分

Answer 1

不定积分SIN^4XCOS^4XDX
=1/16*S（sin2x)^4 dx
=1/64*S(1-cos4x)^2 dx
=1/64*S(1-2cos4x+(cos4x)^2)dx
=1/64*(Sdx-S2cos4xdx+1/2*S(1+cos8x)dx)
=1/64*x-1/128*sin4x+1/128*Sdx+1/128*Scos8xdx
=x/64-1/128*sin4x+x/128+1/1024*sin8x+c
=3x/128-1/128*sin4x+1/1024*sin8x+c

Answer 2

∫SIN^4XCOS^4XDX
=∫sin^2(2x)/4dx
=∫(sin^2(2x)-1/2+1/2)/4dx
=∫(1-(1-2sin^2(2x))/8dx
=∫(1-cos(4x))/8dx
=x/8-sin(4x)/32+C