求积分：∫(sin(a))^6da

∫(sin(a))^6da=
∫(-1/8)*(cos(2a)-1)^3da=
∫((1/8)-(3/8)*cos(2a)+(3/8)*cos(2a)^2-(1/8)*cos(2a)^3)da=
(1/8)a-(3/8)∫cos(2a)da+!##$@#%$#^%^%$^
........
(疯了,步骤实在太长太长了.....)#%$^**(&^*%^$&%$^$#%
........
直接告诉你结果吧:
=(5/16)a-(1/4)sin(2a)+(3/64)sin(4a)+(1/48)sin(2a)^3

-(1/6)*sin(a)^5*cos(a)-(5/24)*sin(a)^3*cos(a)-(5/16)*cos(a)*sin(a)+(5/16)*a

先二倍角公式降次，得cos2a的三次多项式

常数，一次容易，二次的再用一次二倍角，三次的把一个弄到d里面“凑微分”

(60 x - 45 Sin[2 x] + 9 Sin[4 x] - Sin[6 x])/192

∫(sin(a))^6da
=-∫sina^5dcosa
=-sina^5cosa+∫cosa^25sina^4da
=-sina^5cosa+5∫(1-sina^2)sina^4da
6∫(sin(a))^6da=-sina^5cosa+5∫sina^4da
=-sina^5cosa-5∫sina^3dcosa
=-sina^5cosa-5sina^3cosa+5∫cosa^23sina^2da
=-sina^5cosa-5sina^3cosa+15∫1/4sin2a^2da
=-sina^5cosa-5sina^3cosa+15/8∫1-cos4ada
==-sina^5cosa-5sina^3cosa+15/8a-15/32sin4a+C