不定积分习题 ∫sinx^2dx

\u4e0d\u5b9a\u79ef\u5206\u4e60\u9898 \u222bsinxcos2xdx \u222bx (x-1)^1/2dx \u8c01\u77e5\u9053\u8fd9\u4e48\u505a\u554a

\u222bsinxcos2xdx
=\u222bsinx(2cos²x-1)dx
=-\u222b(2cos²x-1)d(cosx)
=-(2/3)cos³x+cosx+C

\u222bx (x-1)^1/2dx
\u4ee4(x-1)^1/2=u\uff0cx=u²+1\uff0cdx=2udu
=\u222b(u²+1)u*2udu
=2\u222b(u^4+u²)du
=(2/5)u^5+(2/3)u³+C
=(2/5)(x-1)^(5/2)+(2/3)(x-1)^(3/2)+C

∫sin²xdx=∫(1-cos2x)/2dx=(1/2)∫(1-cos2x)dx=(1/2)(x-∫cos2xdx)=x/2-(1/2)(sin2x)/2+C=x/2-(sin2x)/4+C

原试=∫1/2sinx^2dx^2
=-1/2cosx^2