∫+1-cos2xdx
@晋剂4025:求定积分∫根号1+cos2xdx,积分上限是π,积分下限是0的值? -
归毅18889581820…… ∫根号1+cos2xdx,积分上限是π,积分下限是0的值是2√ ∵cos2x=2cos²x-1 ∴∫√(1+cos2x)dx=∫√2|cosx|dx ∴(0,π)∫√(1+cos2x)dx =(0,π/2)∫√2cosxdx+(π/2,π)∫-√2cosxdx =2√2 所以∫根号1+cos2xdx,积分上限是π,积分下限是0的值是2√2. ...
@晋剂4025:不定积分计算:∫(x - 1)cos2xdx -
归毅18889581820…… ∫(x-1)cos2xdx=∫(xcos2x-cos2x)dx=∫xcos2xdx-∫cos2xdx=1/2∫xdsin2x-∫cos2xdx=1/2(xsin2x-∫sin2xdx)(分部积分法)-∫cos2xdx=1/2xsin2x-1/2∫sin2xdx-∫cos2xdx=1/2xsin2x-1/4∫sin2xd2x-∫cos2xdx=1/2xsin2x+1/4cos2x-1/2sin2x 你是第三步错了,不懂的话再问我
@晋剂4025:∫1/1+cos2xdx= -
归毅18889581820…… 1+cos2x=2(cosx)^2 =∫1/2(cosx)^2 dx =0.5 *∫1/(cosx)^2 dx =0.5tanx +C,C为常数
@晋剂4025:∫π01+cos2xdx -
归毅18889581820…… ∫π01+cos2xdx =∫π02cos2xdx =2(∫π20cosxdx-∫ππ2cosxdx) =2(sinx|π20-sinx|ππ2)=22
@晋剂4025:∫cosxsin3x1+cos2xdx. - 作业帮
归毅18889581820…… [答案] 原式=∫ cosx(1-cos2x)d(cosx) 1+cos2x(令t=cosx) =∫ t(t2-1)dt 1+t2 =∫(t- 2t 1+t2)dt = 1 2t2-ln(1+t2)+C = 1 2cos2x-ln(1+cos2x)+C
@晋剂4025:∫1/1+cos2xdx= - 作业帮
归毅18889581820…… [答案] 1+cos2x=2(cosx)^2 =∫1/2(cosx)^2 dx =0.5 *∫1/(cosx)^2 dx =0.5tanx +C,C为常数
@晋剂4025:高数积分如题 -
归毅18889581820…… =∫(1-cos²x)/cosxdx=∫(secx-cosx)dx=ln|secx+tanx|-sinx+C
@晋剂4025:求∫1+cos^x/1+cos2xdx 的不定积分 - 作业帮
归毅18889581820…… [答案] ∫(1+cos²x)/(1+cos2x)dx =∫(1+cos²x)/(2cos²x)dx =1/2*∫dtanx+1/2*∫dx =(tanx+x)/2+C
@晋剂4025:∫cos^ 2Xdx
归毅18889581820…… ∫cos^ 2Xdx =∫(1+cos2X)/2 dx=1/2X + 1/4sin2x + C
@晋剂4025:(x^2 - 1)cos2xdx,求不定积分 -
归毅18889581820…… (x^2-1)cos2xdx =∫(x²-1)cos2xdx =(1/2)∫(x²-1)d(sin2x) =(1/2)[(x²-1)sin2x-∫2xsin2xdx] =(1/2)[(x²-1)sin2x+∫xd(cos2x)] =(1/2)[(x²-1)sin2x+xcos2x-∫cos2xdx] =(1/2)[(x²-1)sin2x+xcos2x-(1/2)sin2x]+C =(1/2)[(x²-3/2)sin2x+xcos2x]+C
归毅18889581820…… ∫根号1+cos2xdx,积分上限是π,积分下限是0的值是2√ ∵cos2x=2cos²x-1 ∴∫√(1+cos2x)dx=∫√2|cosx|dx ∴(0,π)∫√(1+cos2x)dx =(0,π/2)∫√2cosxdx+(π/2,π)∫-√2cosxdx =2√2 所以∫根号1+cos2xdx,积分上限是π,积分下限是0的值是2√2. ...
@晋剂4025:不定积分计算:∫(x - 1)cos2xdx -
归毅18889581820…… ∫(x-1)cos2xdx=∫(xcos2x-cos2x)dx=∫xcos2xdx-∫cos2xdx=1/2∫xdsin2x-∫cos2xdx=1/2(xsin2x-∫sin2xdx)(分部积分法)-∫cos2xdx=1/2xsin2x-1/2∫sin2xdx-∫cos2xdx=1/2xsin2x-1/4∫sin2xd2x-∫cos2xdx=1/2xsin2x+1/4cos2x-1/2sin2x 你是第三步错了,不懂的话再问我
@晋剂4025:∫1/1+cos2xdx= -
归毅18889581820…… 1+cos2x=2(cosx)^2 =∫1/2(cosx)^2 dx =0.5 *∫1/(cosx)^2 dx =0.5tanx +C,C为常数
@晋剂4025:∫π01+cos2xdx -
归毅18889581820…… ∫π01+cos2xdx =∫π02cos2xdx =2(∫π20cosxdx-∫ππ2cosxdx) =2(sinx|π20-sinx|ππ2)=22
@晋剂4025:∫cosxsin3x1+cos2xdx. - 作业帮
归毅18889581820…… [答案] 原式=∫ cosx(1-cos2x)d(cosx) 1+cos2x(令t=cosx) =∫ t(t2-1)dt 1+t2 =∫(t- 2t 1+t2)dt = 1 2t2-ln(1+t2)+C = 1 2cos2x-ln(1+cos2x)+C
@晋剂4025:∫1/1+cos2xdx= - 作业帮
归毅18889581820…… [答案] 1+cos2x=2(cosx)^2 =∫1/2(cosx)^2 dx =0.5 *∫1/(cosx)^2 dx =0.5tanx +C,C为常数
@晋剂4025:高数积分如题 -
归毅18889581820…… =∫(1-cos²x)/cosxdx=∫(secx-cosx)dx=ln|secx+tanx|-sinx+C
@晋剂4025:求∫1+cos^x/1+cos2xdx 的不定积分 - 作业帮
归毅18889581820…… [答案] ∫(1+cos²x)/(1+cos2x)dx =∫(1+cos²x)/(2cos²x)dx =1/2*∫dtanx+1/2*∫dx =(tanx+x)/2+C
@晋剂4025:∫cos^ 2Xdx
归毅18889581820…… ∫cos^ 2Xdx =∫(1+cos2X)/2 dx=1/2X + 1/4sin2x + C
@晋剂4025:(x^2 - 1)cos2xdx,求不定积分 -
归毅18889581820…… (x^2-1)cos2xdx =∫(x²-1)cos2xdx =(1/2)∫(x²-1)d(sin2x) =(1/2)[(x²-1)sin2x-∫2xsin2xdx] =(1/2)[(x²-1)sin2x+∫xd(cos2x)] =(1/2)[(x²-1)sin2x+xcos2x-∫cos2xdx] =(1/2)[(x²-1)sin2x+xcos2x-(1/2)sin2x]+C =(1/2)[(x²-3/2)sin2x+xcos2x]+C