(1-x2)3/2次方的原函数,怎么求?

令x=sint,则dx=costdt
∫√(1-x²)³dx
=∫cos³t×costdt
=∫(cos²t)²dt
=∫[(1+cos2t)/2]²dt
=(1/4)∫(cos²2t+2cos2t+1)dt
=(1/4)∫cos²2tdt+(1/2)∫cos2tdt+(1/4)∫dt
=(1/8)∫(1+cos4t)dt+(1/4)sin2t+(1/4)t
=(1/32)sin4t+(1/4)sin2t+(3/8)t+C
=(1/8)x|1-2x²|√(1-x²)+(1/2)x√(1-x²)+(3/8)arcsinx+C

我知道