怎么破～…～～…～…～区间[0,2π]的定积分∫(sint-sin2t)²dt 请问sint的2次方与 sin(t^2) 的不定积分一样吗？
(sint^2/t)dt\u7684\u5b9a\u79ef\u5206\u5177\u4f53\u56de\u7b54\u5982\u56fe\u6240\u793a\uff1a
\u4e00\u4e2a\u51fd\u6570\uff0c\u53ef\u4ee5\u5b58\u5728\u4e0d\u5b9a\u79ef\u5206\uff0c\u800c\u4e0d\u5b58\u5728\u5b9a\u79ef\u5206\uff1b\u4e5f\u53ef\u4ee5\u5b58\u5728\u5b9a\u79ef\u5206\uff0c\u800c\u4e0d\u5b58\u5728\u4e0d\u5b9a\u79ef\u5206\u3002\u4e00\u4e2a\u8fde\u7eed\u51fd\u6570\uff0c\u4e00\u5b9a\u5b58\u5728\u5b9a\u79ef\u5206\u548c\u4e0d\u5b9a\u79ef\u5206\uff1b\u82e5\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\uff0c\u5219\u5b9a\u79ef\u5206\u5b58\u5728\uff1b\u82e5\u6709\u8df3\u8dc3\u95f4\u65ad\u70b9\uff0c\u5219\u539f\u51fd\u6570\u4e00\u5b9a\u4e0d\u5b58\u5728\uff0c\u5373\u4e0d\u5b9a\u79ef\u5206\u4e00\u5b9a\u4e0d\u5b58\u5728\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u542b\u6709\u221a\uff08a+bx\uff09\u7684\u79ef\u5206\u516c\u5f0f\u4e3b\u8981\u5305\u542b\u6709\u4ee5\u4e0b\u51e0\u7c7b\uff1a
\u76f4\u89c2\u5730\u8bf4\uff0c\u5bf9\u4e8e\u4e00\u4e2a\u7ed9\u5b9a\u7684\u5b9e\u51fd\u6570f\uff08x\uff09\uff0c\u5728\u533a\u95f4[a\uff0cb]\u4e0a\u7684\u5b9a\u79ef\u5206\u3002
\u82e5f\uff08x\uff09\u5728[a,b]\u4e0a\u6052\u4e3a\u6b63\uff0c\u53ef\u4ee5\u5c06\u5b9a\u79ef\u5206\u7406\u89e3\u4e3a\u5728Oxy\u5750\u6807\u5e73\u9762\u4e0a\uff0c\u7531\u66f2\u7ebf\uff08x\uff0cf\uff08x\uff09\uff09\u3001\u76f4\u7ebfx=a\u3001x=b\u4ee5\u53cax\u8f74\u56f4\u6210\u7684\u9762\u79ef\u503c\uff08\u4e00\u79cd\u786e\u5b9a\u7684\u5b9e\u6570\u503c\uff09\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1\u2014\u2014\u79ef\u5206\u516c\u5f0f
\u4e0d\u4e00\u6837\uff01
\u222b(sint)^2dt=\u222b(1-cos2t)/2dt=1/2-1/2\u222bcos2tdt=1/2-1/4sin2t+C
\u222bsin(t^2)dt\u4ee4x=t^2,\u5219t=x^(1/2)
\u222bsin(t^2)dt=\u222bsinxdx^(1/2)=1/2x^(-1/2)\u222bsinxdx=-1/2x^(-1/2)cosx+C=-1/2x^(-t)cos(t^2)+C
展开，这个不算复杂，不知道你是否了解 点火公式，有这个运算起来比较快捷
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