两道高数题,求过程

2024-12-03 05:11:47
推荐回答(2个)
回答1:

回答2:

①=lim(x→0)[(sec²x-1)/(2xsinxcos3x+x²cosxcos3x-3x²sinxsin3x)] 0/0 洛必达
=lim(x→0)[(tan²x)/(2xsinxcos3x+x²cosxcos3x-3x²sinxsin3x)]
=lim(x→0)[1/(2cos3x+cosxcos3x-3sinxsin3x)] x→0 x~sinx~tanx
=⅓
②=∫d(tant)/(1+tan²t)^1.5 令x=tant
=∫sec²tdt/sec³t
=∫costdt
=sint+C
=x/√(1+x²)+C