搜搜做题作业网sin^2x*cos^4x 不定积分
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/27 16:27:24
sin^2(x)cos^4(x)=1/4*sin²2xcos²x=1/4*(1-cos4x)/2*(1+cos2x)/2=1/16*(1+cos2x-cos4x-cos2xcos4
f(x)=cos^4x－2sinxcosx－sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)-sin2x=cos2x-sin2x=根号2*cos(2x+л/4)(1)f(x)
请采纳再答：
1就是立方和公式.设sin²x=m,cos²x=n左边就是m³+n³=(m+n)(m²-mn+n²)2.cos2x=2cos²x-
求采纳.再问：图不太清楚但谢谢啦😊
原式=-∫cos²xdcosx=-cos³x/3+C再问：第一步能讲一下为什么吗？再答：dcosx=-sinxdx采纳吧
sin^2x+cos^2x=1
=sin^2(x)*[cos^2(x)-1]=-sin^4(x)再答：别忘了负号再问：嗯谢谢
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x
∫cos2xdx/(sin^2xcos^2x)=4∫cos2xdx/(2sinxcosx)^2=4∫cos2xdx/(sin2x)^2=2∫cos2xd(2x)/(sin2x)^2=2∫d(sin2x
2cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1-2cosx^2=sin2x-cos2x=√2sin(2x-π/4)
证明：∵cos²x-sin²x=cos2xcos⁴x+sin⁴x=1-2cos²xsin²x=1-(1-cos4x)/4=3/4+(co
sin^2(x)+cos^2(x+30)+sin(x)cos(x+30）=sin^2(x)+cos(x+30)[cos(x+30)+sinx]=sin^2(x)+cos(x+30)(cosxcos30
sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)
原式=∫dx/(cos²x(1+4tan²x))=∫d(tanx)/(1+4tan²x)=1/2∫d(2tanx)/(1+(2tanx)²)=arctan(2t
sin^4x-sin^2x+cos^2x=sin^2x*(sin^2x-1)+cos^2x=-sin^2x*cos^2x+cos^2x=cos^2x*(1-sin^2x)=cos^2x*cos^2x=
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x
先改写被积函数如图,再用凑微分法计算.经济数学团队帮你解答,请及时采纳.再问：再问：请问这个不定积分怎么求？再答：之前不是有人给你答过了吗？用代换x+(1/2)=(1/2)secu，不过计算比较烦，我