∫cos^4xsin^2xdx怎么积分?
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/02 16:26:03
∫cos^4xsin^2xdx怎么积分?
∵cos^4xsin^2x
=cos^4x(1-cos²x)
=cos^4x-cos^6x
=[1+cos(2x)]²/4-[1+cos(2x)]³/8
=[1+2cos(2x)+cos²(2x)]/4-[1+3cos(2x)+3cos²(2x)+cos³(2x)]/8
=1/8+1/8cos(2x)-1/8cos²(2x)-1/8cos³(2x)
=1/8+1/8cos(2x)-[1+cos(4x)]/16-1/8cos³(2x)
=1/16+1/8cos(2x)-cos(4x)/16-1/8cos³(2x)
∴∫cos^4xsin^2xdx
=∫[1/16+1/8cos(2x)-cos(4x)/16-1/8cos³(2x)]dx
=x/16+sin(2x)/16-sin(4x)/64-1/16∫[1-sin²(2x)]d[sin(2x)]
=x/16+sin(2x)/16-sin(4x)/64-[sin(2x)-sin³(2x)/3]/16+C
=x/16-sin(4x)/64+sin³(2x)/48+C, (C是积分常数).
=cos^4x(1-cos²x)
=cos^4x-cos^6x
=[1+cos(2x)]²/4-[1+cos(2x)]³/8
=[1+2cos(2x)+cos²(2x)]/4-[1+3cos(2x)+3cos²(2x)+cos³(2x)]/8
=1/8+1/8cos(2x)-1/8cos²(2x)-1/8cos³(2x)
=1/8+1/8cos(2x)-[1+cos(4x)]/16-1/8cos³(2x)
=1/16+1/8cos(2x)-cos(4x)/16-1/8cos³(2x)
∴∫cos^4xsin^2xdx
=∫[1/16+1/8cos(2x)-cos(4x)/16-1/8cos³(2x)]dx
=x/16+sin(2x)/16-sin(4x)/64-1/16∫[1-sin²(2x)]d[sin(2x)]
=x/16+sin(2x)/16-sin(4x)/64-[sin(2x)-sin³(2x)/3]/16+C
=x/16-sin(4x)/64+sin³(2x)/48+C, (C是积分常数).
求下列函数积分1)∫xsin^2xdx
求下列不定积分:1、(cot)^2•xdx 2、cos2x/(cos^2xsin^2x)dx
∫ cos²xsin²x dx求积分步骤
怎么计算定积分 ∫(0,4)2/1+√xdx?
求积分 cos2x/cos^xsin^2x dx
求定积分∫上限π下限0 cos xdx
计算定积分:∫(0,π) cos²xdx
数学 积分sin^3x/cos^4xdx 积分cos^4x/sin^3xdx 请详细解答,谢谢!
求定积分(x³+sin²x)cos^4 xdx 范围为(-π/2,π/2)
定积分(兀/2到-兀/2)根号cos^3x-cos^5xdx
计算下列不定积分(1)∫xe^(-3x)dx(2)∫xcos(4x+3)dx(3)∫xsin^2 xdx(4)∫x^2
∫cos根号x/根号xdx 用换元积分法求的~