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∫(lntAnx/sin2x)Dx

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∫(1+tanx)/sin2x dx =∫ ( 1+ sinx/cosx) /( 2sinx.cosx) dx =∫ ( cosx+ sinx ) /[ 2sinx.(cosx)^2 ] dx =∫ dx/( 2sinx.cosx) +(1/2)∫ (secx)^2 dx =∫ dx/sin2x +(1/2)tanx =∫ csc2x dx +(1/2)tanx =(1/2)ln|csc2x -cot2x| +(1/2)tanx + C

见图

y=Asin(ωx α)和y=Acos(ωx α)的最小正周期T=2π/ω.y=tan(ωx α)和y=cot(ωx α)的最小正周期T=π/ω.其中ω是x的系数,也叫三角函数的圆频率。sin2x的ω=2,故T=2π/2=π.tan(x/2)的ω=1/2,故T=π/(1/2)=2π.下面求f(x)=sin2x tan(x/2)的最小正周期。ω1=2,...

x趋于0吧 limtanx^sin2x=lime^(lntanx*sin2x) limlntanx*sin2x=lim2xlntanx 接下来用洛必达法则 最后结果为1

设 t=sinx-cosx= 2 sin(x- π 4 ) ,…(2分)则t ∈[- 2 , 2 ] ,sin2x=2sinxcosx=1-t 2 .…(6分)∴y=sin2x-2sinx+2cosx=1-t 2 -2t=-(t+1) 2 +2.…(8分)∴当t= 2 时,即x=2k π+ 3π 4 ,k∈Z时,y取得最小值为 -1-2 2 ;…(11分)当t=1时,即x...

设tan(x/2)=t,则x=2arctan(t) =∫[3(n^2+1)/(2n^2+4) * 2/(n^2+1)]dx + ln(3+cosx) + C =∫3/(2+n^2)dx + ln(3+cosx) + C =(3/2)∫1/1+(n/√2)^2 + ln(3+cosx) + C =(3/2)arctan[tan(x/2)/√2] + ln(3+cosx) + C

=sin2xd2x/(sin^2 2x) =-dcos2x/(1-cos^2 2x) =-1/2dcos2x/(1-cos2x)-1/2dcos2x/(1+cos2x) =1/2dln(1-cos2x)-1/2dln(1+cos2x) =1/2dln[(1-cos2x)/(1+cos2x)] =dlntanx or =dx/sinxcosx =sec^xdx/tanx =dtanx/tanx =dlntanx

y=sin2x+sinx+cosx+2 y=2sinxcosx+1+sinx+cosx+1 y=[2sinxcosx+sin^2(x)+cos^2(x)]+sinx+cosx+1 y=(sinx+cosx)^2+(sinx+cosx)+1 y=(sinx+cosx+1/2)^2+3/4 又因为:负根号2

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