1、由于被积函数是奇函数,积分区间是对称区间,因此结果为0.
2、∫[0--->a] (a²-x²)^(5/2) dx
换元法：令x=asint,(a²-x²)^(5/2)=a⁵cos⁵t,dx=acostdt,t：0---->π/2
=∫[0--->π/2] a⁵cos⁵t*acost dt
=a⁶∫[0--->π/2] cos⁶t dt
降幂,以下省略积分限
=(a⁶/8)∫(1+cos2t)³dt
=(a⁶/8)∫(1+3cos2t+3cos²2t+cos³2t)dt
里面的cos²2t还需要降幂
=(a⁶/8)∫(1+3cos2t+(3/2)(1+cos4t)+cos³2t)dt
=(a⁶/8)∫(5/2+3cos2t+(3/2)cos4t)dt+(1/8)∫cos³2tdt
=(a⁶/8)(5t/2+(3/2)sin2t+(3/8)sin4t)+(1/16)∫cos²2td(sin2t)
=(a⁶/8)(5t/2+(3/2)sin2t+(3/8)sin4t)+(1/16)∫[1-sin²2t]d(sin2t)
=(a⁶/8)(5t/2+(3/2)sin2t+(3/8)sin4t)+(1/16)[sin2t-(1/3)sin³2t] |[0---->π/2]
=5πa⁶/32