Question

正弦函数的4次方的不定积分

要详细步骤，谢谢···

Answer 1

∫(sinx)^4dx
=∫[(sinx)^2]^2dx
=∫1/4(1-cos2x)^2dx
=∫1/4[1-2cos2x+(cos2x)^2]dx
=∫1/4[1-2cos2x+(1+cos4x)/2]dx
=∫(3/8-1/2cos2x+1/8cos4x)dx
=3/8x-1/4sin2x+1/32sin4x+C

Answer 2

In[1]:= Integrate[Sin[x]^4, x]

Out[1]= (3 x)/8 - 1/4 Sin[2 x] + 1/32 Sin[4 x]