∫dx/(sin²x+tan²x)

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/20 18:34:44
∫dx/(sin²x+tan²x)
x){Ա:B_83OMºB$ԴI*'BΆSYYMD @%PAdpQmflJB6PQQ>o WjTa i*2D QWNd-F`a]cL A K/F5 <]ɮ>\XKŋ3o|m:`OvhdgJ W :O;fVFE@0/D8@!@\Ά'"zZ:m#PQ,!vh@ng,P!$فRD

∫dx/(sin²x+tan²x)
∫dx/(sin²x+tan²x)

∫dx/(sin²x+tan²x)
符号难敲
∫dx/(sin²x+tan²x)
=∫[cos²x/(sin²xcos²+sin²x)]dx
=∫[(1+cos2x)/2]/[(1/4)sin²2x+(1-cos2x)/2]dx
=(1/2)∫[(1+cos2x)]/[(1/4)(1-cos²2x+(1-cos2x)/2]dx
=-2∫[(1+cos2x)]/[(cos²2x+2cos2x-3]dx
=(1/2)∫[1/(cost+3)]dt+(1/2)∫[1/(cost-1)]dt
对于∫[1/(cosx+3)]dx这类积分,
万能代换t=tan(x/2),则x=2arctant,dx=2dt/(1+t^2),cosx=(1-t^2)/(1+t^2),所以
∫dx/(cosx+3)=∫dt/(t^2+2)=1/√2×arctan(t/√2)+C=1/√2×arctan(tan(x/2)/√2)+C