求积分∫sincosx sin（cosx）的不定积分怎么求
\u6c42\u89e3\u8fd9\u4e2a\u4e0d\u5b9a\u79ef\u5206\u222bsin^4xcosxdx\uff01\u524d\u9762\u662fsinx\u76844\u6b21\u65b9\u8fd8\u662fsin4x\u554a
\u222bsin(cosx)dx
\u8bbe cosx=t ,x=arccost,dx=-1/(\u221a(1-t²))dt
\u222bsin(cosx)dx
=\u222bsint*(-1)/(\u221a(1-t²))dt \
=-\u222bsint/(\u221a(1-t²))dt
因为公式多，看一下
http://arxiv.org/pdf/0708.3254
的Theorem 5.——application 1.
连dx都没有,算是积分吗?
还有就算加上dx
也是∫sin(cosx)dx,还是∫sinxcoxdx?
∫sinxcosxdx(way1)=∫sinxdsinx=0.5*(sinx)^2+C1
way2=-∫cosxdcosx=0.5*(cosx)^2+C2
way3=∫0.5sin2xdx=0.25∫sin2xd2x=-0.25cos2x+C3
∫sinxcosxdx=∫sinxdsinx=(1/2)sin^2
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