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∫(lntAnx/sin2x)Dx

∫(lntanx/sin2x)dx =∫(lntanx)/2sinxcosx)dx =½∫(lntanx)cosx/(sinxcos²x)dx =½∫(lntanx)cosx/(sinx)dtanx =½∫(lntanx)/tanx)dtanx =½∫(lntanx)d(lntanx) =¼ [ln(tanx)]² + C

∫(lntanx)/(sin2x) dx =(1/2)∫(lntanx)/(sinxcosx) dx d(lntanx)=(sec²x/tanx)dx=dx/(sinxcosx) =(1/2)∫(lntanx)/(sinxcosx)·sinxcosxd(lntanx) =(1/2)∫(lntanx)d(lntanx) =(1/2)(lntanx)²/2+C =(1/4)(lntanx)²+C

∫(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

供参考(图)

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

x→0+时 sin2xlncotx→2x(-lntanx) →-2xlnx→-2lnx/(1/x) →(-2/x)/(-1/x^2)→2x→0, ∴e^(sin2xlncotx)→1.

见图

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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