sinx^4+cosx^2怎么化简 求(sinx)^4(cosx)^2的不定积分
sinx^4+cosx^4\u600e\u4e48\u5316\u7b80?sinx^4+cosx^4=sinx^4+cosx^4+2sinx^2cosx^2-2sinx^2cosx^2
=(sinx^2+cosx^2)^2-2sinx^2cosx^2
=1-2sinx^2cosx^2
=1-0.5*(2sinxcosx)^2
=1-0.5*(sin2x)^2
\u4e0d\u77e5\u9053\u5316\u7b80\u6210\u8fd9\u79cd\u7a0b\u5ea6\u591f\u4e0d\u591f
\u2235(sinx)^4(cosx)^2
=(1-cos2x)^2(1+cos2x)/8
=[1-(cos2x)^2](1-cos2x)/8
=(sin2x)^2(1-cos2x)/8
=[1-(cos4x)]/16-(sin2x)^2(cos2x)/8
\u2234\u539f\u79ef\u5206=\u222b[1-(cos4x)]/16*dx-\u222b(sin2x)^2(cos2x)/8*dx
=x/16-(sin4x)]/64-1/16*\u222b(sin2x)^2(dsin2x)
=x/16-(sin4x)]/64-(sin2x)^3/48+C
\u6ce8\uff1a\u7c7b\u4f3c\u4e0a\u9762\u7684\u79ef\u5206\u90fd\u53ef\u4ee5\u5148\u5c06\u88ab\u79ef\u7684\u51fd\u6570\u5316\u7b80\u964d\u6b21\uff0c\u7136\u540e\u518d\u79ef\u5206\u3002\u8981\u6ce8\u610f\u6bcf\u4e00\u9879\u5e94\u4e3a\u540c\u89d2\u7684\u6b63\u4f59\u5f26\uff0c\u6216\u8005\u53ef\u518d\u964d\u6b21\uff0c\u6216\u8005\u7528\u6362\u5143\u79ef\u5206\u3002\u591a\u505a\u51e0\u9053\u9898\uff0c\u76f8\u4fe1\u4f60\u5c31\u80fd\u638c\u63e1\u4e86\uff0c\u795d\u4f60\u987a\u5229\u3001\u6210\u529f\uff01
所以sinxcosx=-1/4
(sinx^2+cosx^2)^2=sinx^4+cosx^4+2sinx^2cosx^2=sinx^4+cosx^4+2*(-1/4)^2=sinx^4+cosx^4+1/8=1
所以sinx^4+cosx^4=7/8
1+(sinx)^2=sinx^2+cosx^2+sinx^2=2sinx^2+cosx^2=3sinxcosx
两边同时除以cosx^2
2tanx^2+1=3tanx
tanx=?
没计算器在身边 您自己算一下
这样做
sinx^4+cosx^2=sinx^4+1-sinx^2=sinx^2(sinx^2-1)+1=sinx^2(-cosx^2)+1=-1/4sin2x^2+1=-1/4((1-cos4x)/2)+1=-1/8(1-cos4x)+1=1/8(cos4x)+7/8应该是这样
主要思想是降幂
利用sinx^2+cosx^2=1
和
cos2x=1-2sinx^2
其余都是这个变形
题目做多了就知道怎么做了
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